Math for Business and Economics: Compendium of Essential Formulas [2 ed.] 3662669749, 9783662669747, 9783662669754

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Franz W. Peren

Math for Business and Economics Compendium of Essential Formulas 2nd Edition

Math for Business and Economics

Franz W. Peren

Math for Business and Economics Compendium of Essential Formulas Second Edition

Franz W. Peren Bonn-Rhein-Sieg University Sankt Augustin, Germany

ISBN 978-3-662-66974-7 ISBN 978-3-662-66975-4 (eBook) https://doi.org/10.1007/978-3-662-66975-4 © Springer-Verlag GmbH Germany, part of Springer Nature 2021, 2023 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer-Verlag GmbH, DE, part of Springer Nature. The registered company address is: Heidelberger Platz 3, 14197 Berlin, Germany

For my father, Paul.

Preface

Preface to the 2nd , revised and supplemented edition The 2nd edition of this compendium of formulas for math for business and economics has been revised and supplemented partially. My valuable research assistant Nawid Schahab has contributed to the current edition. He deserves my thanks. Should any mistakes remain, such errors shall be exclusively at the expense of the author. The author is thankful in advance to all users of this formulary for any constructive comments or suggestions. Bonn, January 2023

Franz W. Peren

Preface to the 1st edition The following book is based on the author’s expertise in the field of business mathematics. After completing his studies in business administration and mathematics, he started his career working for a global bank and the German government. Later he became a professor of business administration, specialising in quantitative methods. He has been a professor at the Bonn-Rhein-Sieg University in Sankt Augustin, Germany since 1995, where he is mainly teaching business mathematics, business statistics, and operations research. He has also previously taught and conducted research at the University of Victoria in Victoria, BC, Canada and at Columbia University in New York City, New York, USA. To the author’s best knowledge and beliefs, this formulary presents its mathematical contents in a practical manner, as they are needed for meaningful and relevant application in global business, as well as in universities and economic practice. The author would like to thank his academic colleagues who have contributed to this work and to many other projects with creativity, knowledge and dedication for more than 25 years. In particular, he would VII

VIII

Preface

like to thank Ms. Eva Siebertz and Mr. Nawid Schahab, who were instrumental in managing and creating this formulary. Special thanks are given to Ms. Camilla Demuth, Ms. Linh Hoang, and Ms. Michelle Jarsen. Should any mistakes remain, such errors shall be exclusively at the expense of the author. The author is thankful in advance to all users of this formulary for any constructive comments or suggestions. Bonn, March 2021

Franz W. Peren

Contents

List of Abbreviations 1

Mathematical Signs and Symbols

XXI 1

1.1

Pragmatic Signs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

1.2

General Arithmetic Relations and Links . . . . . . . . . . . .

1

1.3

Sets of Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2

1.4

Special Numbers and Links . . . . . . . . . . . . . . . . . . . . . .

3

1.5

Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3

1.6

Exponential Functions, Logarithm . . . . . . . . . . . . . . . .

4

1.7

Trigonometric Functions, Hyperbolic Functions . . . . .

4

1.8

Vectors, Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5

1.9

Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6

1.10

Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7

1.11

Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7

1.12

Order Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7

1.13

SI Multiplying and Dividing Prefixes . . . . . . . . . . . . . . .

8

1.14

Greek Alphabet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

9

IX

X 2

3

Contents Logic

11

2.1

Mathematical Logic . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11

2.2

Propositional Logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Propositional Variable . . . . . . . . . . . . . . . . . . . 2.2.2 Truth Tables . . . . . . . . . . . . . . . . . . . . . . . . . . .

11 11 12

Arithmetic

15

3.1

Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.2 Set Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.3 Set Operations . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.4 Relations, Laws, Rules of Calculation for Sets 3.1.5 Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.6 Numeral Systems . . . . . . . . . . . . . . . . . . . . . . . 3.1.6.1 Decimal System (Decadic System) 3.1.6.2 Dual System (Binary System) . . . . 3.1.6.3 Roman Numeral System . . . . . . . . .

15 15 16 17 19 21 22 23 23 24

3.2

Elementary Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Elementary Foundations . . . . . . . . . . . . . . . . 3.2.1.1 Axioms . . . . . . . . . . . . . . . . . . . . . . . 3.2.1.2 Factorisation . . . . . . . . . . . . . . . . . . . 3.2.1.3 Relations . . . . . . . . . . . . . . . . . . . . . . 3.2.1.4 Absolute Value, Signum . . . . . . . . . 3.2.1.5 Fractions . . . . . . . . . . . . . . . . . . . . . . 3.2.1.6 Polynomial Division . . . . . . . . . . . . . 3.2.1.7 Horner’s Scheme (Horner’s Method) 3.2.2 Conversions of Terms . . . . . . . . . . . . . . . . . . . 3.2.2.1 Binomial Formulas . . . . . . . . . . . . . . 3.2.2.2 Binomial Theorem . . . . . . . . . . . . . . 3.2.2.3 General Binomial Theorem for Natural Exponents . . . . . . . . . . . . . . . . . 3.2.2.4 General Binomial Theorem for Real Exponents . . . . . . . . . . . . . . . . . . . . . 3.2.2.5 Polynomial Terms . . . . . . . . . . . . . . . 3.2.3 Summation and Product Notation . . . . . . . . . 3.2.3.1 Summation Notation . . . . . . . . . . . .

24 24 25 25 26 26 27 27 29 30 30 31 31 31 32 32 32

Contents

XI 3.2.3.2 Product Notation . . . . . . . . . . . . . . . Powers, Roots . . . . . . . . . . . . . . . . . . . . . . . . . . Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Factorial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Binomial Coefficient . . . . . . . . . . . . . . . . . . . . .

33 34 37 39 40

3.3

Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.2 Limit of a Sequence . . . . . . . . . . . . . . . . . . . . 3.3.3 Arithmetic and Geometric Sequences . . . . .

41 41 44 46

3.4

Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.2 Arithmetic and Geometric Series . . . . . . . . . .

47 47 47

3.2.4 3.2.5 3.2.6 3.2.7

4

Algebra

51

4.1

Fundamental Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

51

4.2

Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Linear Equations with One Variable . . . . . . . 4.2.2 Linear Inequations with One Variable . . . . . 4.2.3 Linear Equations with Multiple Variables . . . 4.2.4 Systems of Linear Equations . . . . . . . . . . . . . 4.2.5 Linear Inequations with Multiple Variables .

53 53 56 56 57 61

4.3

Non-linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.1 Quadratic Equations with One Variable . . . . 4.3.2 Cubic Equations with One Variable . . . . . . . 4.3.3 Biquadratic Equations . . . . . . . . . . . . . . . . . . . 4.3.4 Equations of the nth Degree . . . . . . . . . . . . . . 4.3.5 Radical Equations . . . . . . . . . . . . . . . . . . . . . .

62 62 65 67 68 69

4.4

Transcendental Equations . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 Exponential Equations . . . . . . . . . . . . . . . . . . 4.4.2 Logarithmic Equations . . . . . . . . . . . . . . . . . .

71 71 73

4.5

Approximation Methods . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.1 Regula falsi (Secant Method) . . . . . . . . . . . . 4.5.2 Newton’s Method (Tangent Method) . . . . . . .

75 75 77

XII

Contents 4.5.3

5

General Approximation Method (Fixed-point Iteration) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

80

Linear Algebra

87

5.1

Fundamental Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1 Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.2 Equality/Inequality of Matrices . . . . . . . . . . . 5.1.3 Transposed Matrix . . . . . . . . . . . . . . . . . . . . . . 5.1.4 Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.5 Special Matrices and Vectors . . . . . . . . . . . .

87 87 88 89 89 92

5.2

Operations with Matrices . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Addition of Matrices . . . . . . . . . . . . . . . . . . . . . 5.2.2 Multiplication of Matrices . . . . . . . . . . . . . . . . . 5.2.2.1 Multiplication of a Matrix with a Scalar . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.2.2 The Scalar Product of Two Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.2.3 Multiplication of a Matrix by a Column Vector . . . . . . . . . . . . . . . . . . . . 5.2.2.4 Multiplication of a Row Vector by a Matrix . . . . . . . . . . . . . . . . . . . . . . . 5.2.2.5 Multiplication of Two Matrices . . . .

94 94 96

5.3

96 98 100 102 103

The Inverse of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.2 Determination of the Inverse with the Usage of the Gaussian Elimination Method . . . . . .

107 107 109

5.4

The Rank of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.2 Determination of the Rank of a Matrix . . . . .

113 113 113

5.5

The Determinant of a Matrix . . . . . . . . . . . . . . . . . . . . . 5.5.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.2 Calculation of Determinants . . . . . . . . . . . . . . 5.5.3 Characteristics of Determinants . . . . . . . . . . .

117 117 118 124

Contents 5.6

6

7

XIII The Adjoint of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6.2 Determination of the Inverse with the Usage of the Adjoint . . . . . . . . . . . . . . . . . . . . . . . . . . .

125 125 127

Combinatorics

129

6.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

129

6.2

Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

133

6.3

Variations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

135

6.4

Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

136

Financial Mathematics

141

7.1

141 141 142 142

Calculation of Interest . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.1 Fundamental Terms . . . . . . . . . . . . . . . . . . . . . 7.1.2 Annual Interest . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.2.1 Simple Interest Calculation . . . . . . . 7.1.2.2 Compound Computation of Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.2.3 Composite Interest . . . . . . . . . . . . . 7.1.3 Interest During the Period . . . . . . . . . . . . . . . 7.1.3.1 Simple Interest Calculation (linear) 7.1.3.2 Simple Interest Using the Nominal Annual Interest Rate . . . . . . . . . 7.1.3.3 Compound Interest (exponential) . 7.1.3.4 Interest with Compound Interest Using a Conforming Annual Interest Rate . . . . . . . . . . . . . . . . . . . . . . 7.1.3.5 Mixed Interest . . . . . . . . . . . . . . . . . 7.1.3.6 Steady Interest Rate . . . . . . . . . . . .

161 162 163

7.2

Annual Percentage Rate . . . . . . . . . . . . . . . . . . . . . . . .

168

7.3

Depreciation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.1 Time Depreciation . . . . . . . . . . . . . . . . . . . . . . 7.3.1.1 Linear Depreciation . . . . . . . . . . . . .

173 173 173

144 146 158 159 159 160

XIV

Contents 7.3.1.2 7.3.1.3 7.3.2 7.3.3

7.4

7.5

Arithmetic-Degressive Depreciation Geometric-Degressive Depreciation . . . . . . . . . . . . . . . . . . . . . . . . . . . Units of Production Depreciation . . . . . . . . . Extraordinary Depreciation . . . . . . . . . . . . . .

Annuity Calculation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.1 Fundamental Terms . . . . . . . . . . . . . . . . . . . . . 7.4.2 Finite, Regular Annuity . . . . . . . . . . . . . . . . . . 7.4.2.1 Annual Annuity with Annual Interest 7.4.2.2 Annual Annuity with Sub-Annual Interest . . . . . . . . . . . . . . . . . . . . . . . 7.4.2.3 Sub-Annual Annuity with Annual Interest . . . . . . . . . . . . . . . . . . . . . . . 7.4.2.4 Sub-Annual Annuity with Sub-Annual Interest . . . . . . . . . . . . . . . . . . . . . . . 7.4.3 Finite, Variable Annuity . . . . . . . . . . . . . . . . . . 7.4.3.1 Irregular Annuity . . . . . . . . . . . . . . . 7.4.3.2 Arithmetic Progressive Annuity . . . 7.4.3.3 Geometric Progressive Annuity . . . 7.4.4 Perpetuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Sinking Fund Calculation . . . . . . . . . . . . . . . . . . . . . . . . 7.5.1 Fundamental Terms . . . . . . . . . . . . . . . . . . . . . 7.5.2 Annuity Repayment . . . . . . . . . . . . . . . . . . . . . 7.5.3 Repayment by Instalments . . . . . . . . . . . . . . 7.5.4 Repayment with Premium . . . . . . . . . . . . . . . 7.5.4.1 Annuity Repayment with Premium 7.5.4.2 Repayment of an Instalment Debt with Premium . . . . . . . . . . . . . . . . . . 7.5.5 Repayment with Discount (Disagio) . . . . . . . 7.5.5.1 Annuity Repayment with Discount when Immediately Booked as Interest Expense . . . . . . . . . . . . . . . . . . . 7.5.5.2 Annuity Repayment with Discount when a Disagio is Included in Prepaid Expenses . . . . . . . . . . . . . . . . . 7.5.5.3 Instalment Repayment with Discount when Immediately Booked as Interest Expense . . . . . . . . . . . . .

174 176 178 179 180 180 183 183 187 190 194 213 213 220 231 234 235 236 238 241 243 243 248 249

251

253

253

XV

Contents 7.5.5.4

7.5.6 7.5.7

7.5.8

7.6

8

Instalment Repayment with Discount when a Disagio is Included in Prepaid Expenses . . . . . . . . . . . . Grace Periods . . . . . . . . . . . . . . . . . . . . . . . . . Rounded Annuities . . . . . . . . . . . . . . . . . . . . . 7.5.7.1 Percentage Annuity . . . . . . . . . . . . . 7.5.7.2 Repayment of Bonds . . . . . . . . . . . Repayment During the Year . . . . . . . . . . . . . . 7.5.8.1 Annuity Repayment During the Year 7.5.8.2 Repayment by Instalments During the Year . . . . . . . . . . . . . . . . . . . . . . .

Investment Calculation . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.1 Fundamental Terms . . . . . . . . . . . . . . . . . . . . . 7.6.2 Fundamentals of Financial Mathematics . . . 7.6.3 Methods of Static Investment Calculation . . 7.6.4 Methods of Dynamic Investment Calculation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.4.1 Net Present Value Method (Net Present Value, Amount of Capital, Final Asset Value) . . . . . . . 7.6.4.2 Annuity Method . . . . . . . . . . . . . . . . 7.6.4.3 Internal Rate of Return Method . .

254 255 257 257 260 266 266 274 279 280 283 286 286

287 290 293

Optimisation of Linear Models

297

8.1

Lagrange Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.2 Formation of the Lagrange Function . . . . . . . 8.1.3 Determination of the Solution . . . . . . . . . . . . . 8.1.4 Interpretation of λ . . . . . . . . . . . . . . . . . . . . . . .

297 297 297 298 299

8.2

Linear Optimisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.2 The Linear Programming Approach . . . . . . . 8.2.3 Graphical Solution . . . . . . . . . . . . . . . . . . . . . . 8.2.4 Primal Simplex Algorithm . . . . . . . . . . . . . . . . 8.2.5 Simplex Tableau (Basic Structure) . . . . . . . . . 8.2.6 Dual Simplex Algorithm . . . . . . . . . . . . . . . . . .

302 302 303 303 308 308 315

XVI 9

Contents

Functions

325

9.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1.1 Composition of Functions . . . . . . . . . . . . . . . 9.1.2 Inverse Function . . . . . . . . . . . . . . . . . . . . . . .

325 329 331

9.2

Classification of Functions . . . . . . . . . . . . . . . . . . . . . . 9.2.1 Rational Functions . . . . . . . . . . . . . . . . . . . . . 9.2.1.1 Polynomial Functions . . . . . . . . . . . 9.2.1.2 Broken Rational Functions . . . . . . . 9.2.2 Non-rational Functions . . . . . . . . . . . . . . . . . . 9.2.2.1 Power Functions . . . . . . . . . . . . . . . . 9.2.2.2 Root Function . . . . . . . . . . . . . . . . . 9.2.2.3 Transcendental Functions . . . . . . . . 9.2.2.3.1 Exponential Functions . 9.2.2.3.2 Logarithmic Functions . . 9.2.2.4 Trigonometric Functions (Angle Functions/Circular Functions) . . . . . . . . .

333 334 334 334 338 338 341 342 342 348

Characteristics of Real Functions . . . . . . . . . . . . . . . . 9.3.1 Boundedness . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.2 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.2.1 Axial Symmetry . . . . . . . . . . . . . . . . 9.3.2.2 Point Symmetry . . . . . . . . . . . . . . . . 9.3.3 Transformations . . . . . . . . . . . . . . . . . . . . . . . . 9.3.3.1 Vertex Form . . . . . . . . . . . . . . . . . . . 9.3.4 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.5 Infinite Discontinuities . . . . . . . . . . . . . . . . . . . 9.3.6 Removable Discontinuities . . . . . . . . . . . . . . . 9.3.7 Jump Discontinuities . . . . . . . . . . . . . . . . . . . . 9.3.8 Homogeneity . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.9 Periodicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.10 Zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.11 Local Extremes . . . . . . . . . . . . . . . . . . . . . . . . 9.3.12 Monotonicity . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.13 Concavity and Convexity | Inflection Points . 9.3.14 Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.14.1 Horizontal Asymptotes . . . . . . . . . . 9.3.14.2 Vertical Asymptote . . . . . . . . . . . . . 9.3.14.3 Oblique Asymptote . . . . . . . . . . . . . 9.3.14.4 Asymptotic Curve . . . . . . . . . . . . . . .

382 382 384 384 386 389 391 394 394 396 397 398 399 399 400 401 402 404 405 407 408 409

9.3

354

Contents

XVII 9.3.15 9.3.16

9.4

Tangent Lines to a Curve . . . . . . . . . . . . . . . . Normal Lines to a Curve . . . . . . . . . . . . . . . .

410 411

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

412

10 Differential Calculus 10.1

10.2

Differentiation of Functions with One Independent Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.2 First Derivative of Elementary Functions . . . 10.1.3 Derivation Rules . . . . . . . . . . . . . . . . . . . . . . . 10.1.4 Higher Derivations . . . . . . . . . . . . . . . . . . . . . . 10.1.5 Differentiation of Functions with Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.6 Curve Sketching . . . . . . . . . . . . . . . . . . . . . . .

425 425

Differentiation of Functions with More Than One Independent Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

435

10.2.1 10.2.2 10.2.3

10.2.4 10.3

417

Partial Derivatives (1st Order) . . . . . . . . . . . . . Partial Derivatives (2nd Order) . . . . . . . . . . . . Local Extrema of the Function f = f (x, y) . . 10.2.3.1 Relative Extrema without Constraint of the Function f = f (x, y) . . . . . . . 10.2.3.2 Relative Extrema with m Constraints of the Function f = f (x1 , . . . , xn ) with m < n . . . . . . . . . . . . . . . . . . . . . Differentials of the Function f = f (x1 , ..., xn )

Theorems of Differentiable Functions . . . . . . . . . . . . . 10.3.1 Mean Value Theorem for Differential Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3.2 Generalized Mean Value Theorem for Differential Calculus . . . . . . . . . . . . . . . . . . . . . . . 10.3.3 Rolle’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 10.3.4 L’Hospital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . 10.3.5 Bounds Theorem for Differential Calculus . .

417 417 420 422 424

435 438 440 440

449 453 455 455 456 456 457 458

XVIII

Contents

11 Integral Calculus

459

11.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

459

11.2

The Indefinite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.1 Definition/Determining the Antiderivative . . . 11.2.2 Elementary Calculation Rules for the Indefinite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . .

460 460

11.3

463

The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3.2 Relationship between the Definite and the Indefinite Integral . . . . . . . . . . . . . . . . . . . . . . . 11.3.3 Special Techniques of Integration . . . . . . . . . 11.3.3.1 Partial Integration . . . . . . . . . . . . . . 11.3.3.2 Integration by Substitution . . . . . . .

468 473 473 475

11.4

Multiple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

476

11.5

Integral Calculus and Economic Problems . . . . . . . . . 11.5.1 Cost Functions . . . . . . . . . . . . . . . . . . . . . . . . . 11.5.2 Revenue Function (= Sales Function) . . . . . . 11.5.3 Profit Functions . . . . . . . . . . . . . . . . . . . . . . . . .

477 477 479 480

12 Elasticities

464 464

485

12.1

Definition of Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . .

485

12.2

Arc Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

486

12.3

Point Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

491

12.4

Price Elasticity of Demand εxp . . . . . . . . . . . . . . . . . . .

494

12.5

Cross Elasticity of Demand εxA pB . . . . . . . . . . . . . . . . .

499

12.6

Income Elasticity of Demand εxy . . . . . . . . . . . . . . . . . .

501

13 Economic Functions

503

Contents

XIX

13.1

Supply Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

503

13.2

Demand Function / Inverse Demand Function . . . . . .

505

13.3

Market Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

507

13.4

Buyer’s Market and Seller’s Market . . . . . . . . . . . . . . .

508

13.5

Supply Gap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

509

13.6

Demand Gap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

509

13.7

Revenue Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

511

13.8

Cost Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

517

13.9

Neoclassical Cost Function . . . . . . . . . . . . . . . . . . . . . .

525

13.10 Cost Function According to the Law of Diminishing Returns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

533

13.11 Direct Costs versus Indirect Costs . . . . . . . . . . . . . . . . 13.11.1 One-Dimensional Cost Allocation Principles 13.11.2 Multi-Dimensional Cost Allocation Principles

546 549 551

13.12 Profit Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

554

14 The Peren Theorem The Mathematical Frame in Which We Live

563

A Financial Mathematical Factors

571

B Bibliography

617

Index

625

List of Abbreviations ACT

Advance Corporation Tax

a.m.

ante meridiem

APR

Annual Percentage Rate

AU

area unit(s)

BCD

Binary Coded Decimal

BEP

break-even point

BGB

German Civil Code

bit

binary digit

calcul.

calculate, calculation

cf.

confer

CFPB

Consumer Financial Protection Bureau

cm

centimetre(s)

CM

Contribution Margin

const.

constant

c.p.

ceteris paribus

det

determinant

e

Euler’s number

EC

European Commission

ED

edition

e.g.

exempli gratia

XXI

List of Abbreviations

XXII

etc.

et cetera

EU

European Union

gal.

gallon

gen.

general

h

hour(s)

ICMA

International Capital Market Association

i.e.

id est

incl.

includes, including

inf

infimum

int

integer function

IP

inflection point(s)

IRR

internal rate of return

ISDA

International Swaps and Derivatives Association

j

relative periodic interest rate

kbyte

kilobyte

kg

kilogram(s)

l

litre(s)

lb

pound(s)

lim

limit

ln

natural logarithm

log

logarithm

ltd.

limited

XXIII

List of Abbreviations

LU

length unit(s)

max

maximum, maximise

mbyte

megabyte

min

minimise, minimum

min(s)

minute(s)

mm

millimetre(s)

norm

normal

NPV

net present value

opt

optimise, optimisation

oz

ounce(s)

p.a.

per annum

p.m.

post meridiem

QU

quantity unit(s)

rad

radius

regen

regeneration

rep.

repetition

resp.

respectively

sgn

signum

SI

Système International d’Unités

SS

solution set

SU

universal set

List of Abbreviations

XXIV sup

supremum

TU

time unit(s)

USP

unique selling proposition

U.S.

United States of America

w/

with

w/o

without

yd

yard(s)

Chapter 1

Mathematical Signs and Symbols Remark:

The signs and symbols are partly shown in applications. For definitions see dedicated passage.

1.1 Pragmatic Signs a ≈ b

a approximately similar to b

a≪b

a small towards b, a can be neglected compared to b

a≫b

a large towards b

b b a =

b 10 mm; 1 inch = b 25.4 mm a equivalent to b, e.g. 1 cm =

a ∧ b

a and b

a ∨ b

a or b

...

and so forth (until), omission

1.2 General Arithmetic Relations and Links (a, b are figures, elements, objects) a = b

a equals b, arithmetic fundamental term, identity

a ̸= b

a unequal to b, no identity

a := b

a equals b by definition

a < b

a less than b, fundamental term, e.g. −6 < −2

a > b

a greater than b, e.g. 3 > −8

© Springer-Verlag GmbH Germany, part of Springer Nature 2023 F. W. Peren, Math for Business and Economics, https://doi.org/10.1007/978-3-662-66975-4_1

1

1 Mathematical Signs and Symbols

2 a ≤ b

a less than or (at most) equal to b, a ≤ 8 is equivalent to ] − ∞, 8]

a ≥ b

a greater than or (at least) equal to b, is equivalent to b ≤ a

a + b

a plus b, sum of a and b, arithmetic fundamental term

a − b

a minus b, difference between a and b, single-digit linking sign

a · b

a times b, product of a and b, arithmetic fundamental term

a b n

a divided by b, quotient of a and b, e.g.

∑ ai

Sum over ai of i equals 1 up to n,

i=1

16 4

= 16 ÷ 4 = 4

n

∑ ai = a1 + a2 + a3 + ... + an i=1 n

∏ ai

n

Product over ai of i equals 1 up to n, ∏ ai = a1 · a2 · ... · an

i=1

i=1

1.3 Sets of Numbers N

set of natural numbers, N = {0, 1, 2, ...}

N∗

set of positive natural numbers, N∗ = N \{0} = {1, 2, 3, ...}

Z

set of integers, Z = {... − 2, −1, 0, 1, 2, ...}

Q

set of rational numbers, Q = { ab | a, b ∈ Z, b ̸= 0}

Q∗

set of rational numbers which vary from zero, Q∗ = Q \{0}

Q+

set of positive rational numbers

Q+ 0

set of positive rational numbers plus zero

R

set of real numbers

R∗

set of real numbers which vary from zero

R+

set of positive real numbers

R+ 0

set of positive real numbers plus zero

C

set of complex numbers

]a, b[

open interval from a to b {x | a < x < b}

1.5 Limit

3

]a, ∞[

open, unbounded interval starting at a, {x | a < x}

[a, b]

closed interval from a to b, {x | a ≤ x ≤ b}

[a, ∞[

closed, unbounded interval starting at a, {x | a ≤ x}

[a, b[

left-closed, right-open interval from a to b, {x | a ≤ x < b}

1.4 Special Numbers and Links (a, b ∈ R; n, m ∈ Z; s ∈ N) an √ 1 a = a2 = b √ n

a to the power of n, n th power of a for n ≥ 0 root (square root) of a, equivalent to b2 = a for b ≥ 0, a ≥ 0

1 n

a=a =b

n th root of a, equivalent to bn = a for b ≥ 0, a ≥ 0

n!

n factorial, n! = ∏ni=1 ai = 1 · 2 · 3 · ... · n

sgn a

signum of a (algebraic sign), e.g. sgn(−3) = −1

|a|

absolute value of a, e.g. | − 8| = 8

a[i]

a in the i th position; e.g. 5; 6; 7; a[2] = 6

∞

infinity, note: ∞ is not a number

π

3.1415926...

e

Euler’s Number, e = 2.718281

1.5 Limit lim f (x) = a

x→0

a is the limit of the function f (x) for x towards 0, i.e. x x→0 gradually approaches the value 0, the function’s value f (x) converges (limits) towards a

1 Mathematical Signs and Symbols

4 lim f (x) = b

b is the limit of the function f (x) for x towards ∞

lim f (x) = c

c is the limit of the function f (x) for x towards 5

x→∞

x→5

1.6 Exponential Functions, Logarithm ex

exponential function of x, e to the power of x

ln x

natural logarithm of x to base e; loge x = ln x

loga x

logarithm of x to base a; loga x = y ⇔ ay = x with x ; a > 0 and a ̸= 1

log x

common logarithm of x to base 10 log x = lg x = log10 x

lb x

binary (dyadic) logarithm of x to base 2 lb x = log2 x

1.7 Trigonometric Functions, Hyperbolic Functions sin x

sine of x

cos x

cosine of x

tan x

tangent of x

cot x

cotangent of x

sinh x

hyperbolic sine of x

cosh x

hyperbolic cosine of x

tanh x

hyperbolic tangent of x

coth x

hyperbolic cotangent of x

1.8 Vectors, Matrices

5

arcsin y

arc sine of y

arccos y

arc cosine of y

arctan y

arc tangent of y

arccot y

arc cotangent of y

arsinh y

area hyperbolic sine of y

arcosh y

area hyperbolic cosine of y

artanh y

area hyperbolic tangent of y

arcoth y

area hyperbolic cotangent of y

1.8 Vectors, Matrices a, b, x, y, ... → − 0, 0

→ − − → − signs for vectors, also → a , b ,→ x , −y

|a| = a

zero vector, identity element regarding addition of vectors √ absolute value of a, |a| = a · a

< (a, b)

angle between a and b

a⊥b

a orthogonal to b

a×b

a cross b

A, B, ...

signs for matrices



 a11 . . . a1n  . ..   A= .   .. am1 . . . amn A′

= (ai j ) m, n-matrix A element ai j (i th row, j th column)

transposed matrix for A (A′ )′ = A

1 Mathematical Signs and Symbols

6  En×n

1 0 ... 0

  0 1 ...  =.  ..  0 ...



 0  ..  .  1

identity (unit) matrix; diagonal matrix, whose elements of the main diagonals are all 1 and whose remaining elements are all 0

A−1 n×n

inverse matrix for A, A · A−1 = E

r(A)

rank of A, also Rg(A)

1.9 Sets {a1 , ..., an }

set with the elements a1 , ..., an

a∈A

a is element of A

a∈ /A

a is not element of A e.g.: 3 ∈ / {4, 5, 6}

A⊂B

A is proper subset of B, e.g. every element of A also belongs to B, but B contains at least one element that does not belong to A. For example: A ⊂ B if A = {1; 2; 3; 4} and B = {1; 2; 3; 4; 5; 6}

A⊆B

A ist subset of B, e.g. every element of A also belongs to B. This includes A = B. For example: A ⊆ B if A = {1; 2; 3; 4} and B = {1; 2; 3; 4}

A ̸⊂ B

A is not a proper subset of B, e.g. not every element from A also belongs to B and B contains at least one element that does not belong to A. For example: A ̸⊂ B if A = {1; 2; 3; 7} and B = {1; 2; 3; 4; 5; 6}

A∪B

A union B, A or B, includes common elements

A∩B

A intersection B, A and B, includes all occuring

1.12 Order Structures elements A\B

relative complement of A and B, A not B, e. g. {2, 3, 4}{2, 4} = {3}

B¯

complement of B, includes all elements, which are not included in B

0/ = {}

empty set, includes no elements

1.10 Relations (a, b)

(ordered) pair of a and b, also ⟨a; b⟩

A×B

A cross B, cartesian product of A and B, set of all (ordered) pairs from A and B

1.11 Functions f = f (x)

f of x, f is a function dependent on x

D f ; D( f )

domain of f

R f ; R( f )

range of f , codomain of f

f: A → B

f is a transformation of A into B

1.12 Order Structures min X

minimum of X, least element of X

max X

maximum of X, greatest element of X

sup X

supremum of X, least upper bound of X

inf X

infimum of X, greatest lower bound of X

7

1 Mathematical Signs and Symbols

8

1.13 SI1 Multiplying and Dividing Prefixes d

deci

10−1

da

deca

101

c

centi

10−2

h

hecto

102

m

milli

10−3

k

kilo

103

µ

micro

10−6

M

mega

106

n

nano

10−9

G

giga

109

p

pico

10−12

T

tera

1012

f

femto

10−15

P

peta

1015

a

atto

10−18

E

exa

1018

z

zepto

10−21

Z

zetta

1021

y

yocto

10−24

Y

yotta

1024

1

SI is the abbreviation for an international unit system of physical quantities. The SI (Système International d’Unités) defines seven coherent basic units, which can be represented as products of powers. The SI units thus make it possible to define a certain dimension simply by adding one of the prefixes mentioned above, without adding an additional numerical factor. The SI units include the metre (m), the kilogram (kg), the second (s), the ampere (A), the kelvin (K), the mole (mol), and the candela (cd). Derived units, such as the newton (N), can also be formed using the algebraic relations.

1.14 Greek Alphabet

9

1.14 Greek Alphabet Name

Lower Case Letter

Upper Case Letter

alpha

α

A

beta

β

B

gamma

γ

Γ

delta

δ

∆

epsilon

ε

ε

zeta

ζ

Z

eta

η

H

theta

θ

Θ

iota

ι

I

kappa

κ

K

lambda

λ

Λ

mu

µ

M

nu

ν

N

xi

ξ

Ξ

omicron

o

O

pi

π

Π

rho

ρ

P

sigma

σ

Σ

tau

τ

T

upsilon

υ

ϒ

phi

φ

Φ

chi

χ

X

psi

ψ

Ψ

omega

ω

Ω

Chapter 2

Logic 2.1 Mathematical Logic negation

¬ϕ = not ϕ

ϕ ∧ ψ

conjunction

ϕ and ψ

ϕ ∨ ψ

disjunction

ϕ or ψ

ϕ ∨˙ ψ

alternative

either ϕ or ψ, exclusionary or

ϕ⇒ψ

implication

ϕ implies ψ, ψ follows after ϕ,

¬ϕ , ϕ¯

also written as ϕ → ψ ϕ⇔ψ

equivalence

ϕ equivalent to ψ, ϕ is similar to ψ, also written as ϕ ↔ ψ

ϕ ̸⇔ ψ

anticoincidence

negated equivalence, exclusionary either-or

ϕ←ψ

replication

if; if ψ applies then ϕ follows

∀ x

universal quantifier

for all x (applies)

∃ x

existential quantifier

there is (at least) one x for which it applies

2.2 Propositional Logic 2.2.1 Propositional Variable a, b, . . .

are letters or other symbols which can be used as placeholders for statements or truths.

© Springer-Verlag GmbH Germany, part of Springer Nature 2023 F. W. Peren, Math for Business and Economics, https://doi.org/10.1007/978-3-662-66975-4_2

11

2 Logic

12

2.2.2 Truth Tables a

b

¬a

a∧b

a∨b

t

t

f

t

t

t

f

f

f

t

f

t

t

f

t

f

f

t

f

f

with: t = true f = false

Symbol Meaning A

A is a statement that can be true (t) or false ( f ). truth values

t (true); f (false)

Examples: The statement “7 is a prime number” is true, the statement “8 − 3 = 4” is false, “7x + 4 = 25” is only valid when “x = 3”. “3” is called solution. v(A)

v(A) is referred to as the truth value of the statement A. v(A) = 1 means that A is true and v(A) = 0 means that A is false.

¬A

¯ of the statement A is true when A is The negation ¬ A (or A) false and false when A is true.

A∧B

The conjunction A ∧ B is true when both statements are true and false when there is at least one false statement.

A∨B

The disjunction A ∨ B is true when there is at least one true statement, and false when both statements are false.

2.2 Propositional Logic A ⇒B

13

The implication A ⇒ B means: When A is true, B is also true. A is considered as condition (premise) and B as consequence (conclusion). A ⇒ B is only false when a false conclusion is drawn from a true premise.

A ⇔ B The equivalence A ⇔ B means: When A is true, B is also true and vice versa. A ⇔ B is only false when one of the statements is true and the other one is false. ∃

“There is” (e.g.: ∃x ∈ Θ : x2 = 4 means: there is a rational number x with x2 = 4).

∀

“For all” (e.g.: ∀x ∈ Θ : x2 ≥ 0 means: for all rational numbers x with x2 ≥ 0).

Chapter 3

Arithmetic 3.1 Sets 3.1.1 General Notation {a1 , . . . , an }

set with the elements a1 , . . . , an

{x | A(x)}

quantity of all x, to which A(x) applies

0, / also {}

empty set, includes no elements (no elements included)

a∈A

a is element of A, a, b ∈ A ⇔ a ∈ A ∧ b ∈ A

/A a∈

a is not element of A, e.g. 3 ∈ / {4, 5, 6}

A=B

A equals B (set with identical elements, i.e. set equality)

A⊆B

A is improper subset of B, also A ⊂ B

A⊊B

A is proper subset of B, when: A ⊆ B ∧ A ̸= B, proper inclusion relation “included and unequal”

A⊇B

A is superset of B

A∩B

intersection of A and B, A ∩ B = {x | x ∈ A ∧ x ∈ B}

A∪B

set union of A and B, A ∪ B = {x | x ∈ A ∨ x ∈ B}

A\B

relative complement of A and B, / B} (read: A without B) A \ B = {x | x ∈ A ∧ x ∈

A¯

complement set of A, A¯ = G \ A (G is the universal set)

© Springer-Verlag GmbH Germany, part of Springer Nature 2023 F. W. Peren, Math for Business and Economics, https://doi.org/10.1007/978-3-662-66975-4_3

15

3 Arithmetic

16 A×B

product set of A and B, A × B = {(a, b) | a ∈ A ∧ b ∈ B} power set of A; P(A) = {T | T ⊆ A}

P(A)

P(A) is the set of all subsets T of A

Bounds, Limits of a Set A universal set, SU , is bounded upwards (or downwards) if it has at least one upper (or lower) bound B. If both conditions apply, SU is bounded: B ≥ x (B ≤ x)

with

x ∈ SU

infimum:

inf x

greatest lower bound, upper bound

supremum:

sup x

least upper bound, lower bound

3.1.2 Set Relations Inclusion If A is a subset of B (superset), then each ai ∈ A is also ai ∈ B A⊂B⇔B⊃A

with

x∈A⇒x∈B

Equality (Equivalence: “A equals B”) A=B

with

x (x ∈ A ⇔ x ∈ B)

3.1 Sets

3.1.3 Set Operations Union of two sets A ∪ B; disjunction: “A or B” A ∪ B = {x | x ∈ A ∨ x ∈ B}

Intersection of two sets A ∩ B; conjunction: “A and B” A and B are conjunct for: A ∩ B = {x | x ∈ A ∧ x ∈ B} A and B are disjoint for: A ∩ B = 0/

Relative complement of two sets A \ B, “A without B” A \ B = {x | x ∈ A ∧ x ∈ / B}

Symmetric difference of A and B A△B = (A ∪ B) \ (A ∩ B)

17

3 Arithmetic

18 Complement of the set B Set of all elements, which are not included in B B¯ = {x | x ∈ A ∧ x ∈ / B}

Power set of B Set of all subsets of a set B P(B) = {x | x ⊆ B} always valid: 0/ ∈ P(B) and B ∈ P(B)

Product (cartesian) of two sets A × B, “ A cross B ” A × B (product of two sets) is the set of all ordered pairs of elements (a, b) with a ∈ A and b ∈ B A × B = {(a, b) | a ∈ A; b ∈ B}

A × B ̸= B × A

The product set A1 × A2 × ... × An , n ≥ 1, is the set of all ordered k-tuples (x1 , ..., xn ) of the elements x1 of A1 , x2 of A2 , ... , xn of An .

3.1 Sets

19

3.1.4 Relations, Laws, Rules of Calculation for Sets S = universal set

Idempotent law

A∪A = A A∩A = A

Commutative law

A∩B = B∩A A∪B = B∪A

Associative law

(A ∩ B) ∩C = A ∩ (B ∩C) (A ∪ B) ∪C = A ∪ (B ∪C)

Absorption law

A ∩ (A ∪ B) = A A ∪ (A ∩ B) = A

Distributive law

A ∩ (B ∪C) = (A ∩ B) ∪ (A ∩C) A ∪ (B ∩C) = (A ∪ B) ∩ (A ∪C) A ∪ 0/ = A

A ∪ S= S

A ∩ 0/ = 0/

A∩A = A

A \ A = 0/

A \ 0/ = A

A ∩ S= A

20

Product relations (A ∪ B) ×C = (A ×C) ∪ (B ×C) (A ∩ B) ×C = (A ×C) ∩ (B ×C) A × (B ∪C) = (A × B) ∪ (A ×C) A × (B ∩C) = (A × B) ∩ (A ×C) (A \ B) ×C = (A ×C) \ (B ×C) A × (B \C) = (A × B) \ (A ×C) (A × B) ∪ (C × D) = (A ∪C) × (B ∪ D) (A × B) ∩ (C × D) = (A ∩C) × (B ∩ D) A × B = 0/ ⇔ A = 0/ ∨ B = 0/ A ⊆C∧B ⊆ D ⇒ A×B ⊆C×D

3 Arithmetic

3.1 Sets

21

3.1.5 Intervals An interval is a contiguous subset of real numbers, which is limited by two bounds (= boundary points of the number line) a and b, a < b for all a, b ∈ R

•

closed interval

[a, b] = {x | a ≤ x ≤ b}

•

open interval

]a, b[ = {x | a < x < b}

•

half-open intervals

[a, b[ = {x | a ≤ x < b} ]a, b] = {x | a < x ≤ b}

•

infinite (half-open) intervals ∞; −∞ are “improper numbers” in R with −∞ < a; a < ∞ for all a ∈ R [a; ∞[ = {x | a ≤ x} ]a; ∞[ = {x | a < x} ] − ∞; a] = {x | x ≤ a} ] − ∞; a[ = {x | x < a}

3 Arithmetic

22

3.1.6 Numeral Systems

1

decimal

dual/binary

BCD1

octal

hexadecimal

0

0000

0000 0000

0

0

1

0001

0000 0001

1

1

2

0010

0000 0010

2

2

3

0011

0000 0011

3

3

4

0100

0000 0100

4

4

5

0101

0000 0101

5

5

6

0110

0000 0110

6

6

7

0111

0000 0111

7

7

8

1000

0000 1000

10

8

9

1001

0000 1001

11

9

10

1010

0001 0000

12

A

11

1011

0001 0001

13

B

12

1100

0001 0010

14

C

13

1101

0001 0011

15

D

14

1110

0001 0100

16

E

15

1111

0001 0101

17

F

16

10000

0001 0110

20

10

17

10001

0001 0111

21

11

18

10010

0001 1000

22

12

19

10011

0001 1001

23

13

20

10100

0010 0000

24

14

etc.

etc.

etc.

etc.

etc.

BCD (“binary-coded decimal”) reads the pseudo-decimal numbers. The octal system uses the base 8, the hexadecimal system the base 16.

3.1 Sets

23

3.1.6.1 Decimal System (Decadic System) 10k , k ∈ Z

Decimal powers:

100 = 1 101 = 10

10−1 = 0.1

102 = 100

10−2 = 0.01

etc. Decimal notation of an integer b (k, n ∈ N) n

b = ± ∑ bk · 10k = ± b0 100 + b1 101 + b2 102 + ... + bn−1 10n−1 + bn 10n k=0

Base figures bk ∈ {0, 1, 2, ..., 9}

3.1.6.2 Dual System (Binary System) 1 bit (“binary digit”) symbolizes a “yes - no” decision.

1 byte

=

8 bit

1 kbyte

=

210 byte

=

1.024 byte

=

210

=

1.024 kbyte

1 mbyte

kbyte

etc. Base symbols: 0.1 Place value: powers of 2 n

∑ k=−∞

ak · 2k ak = 0.1




3 Arithmetic

24 3.1.6.3 Roman Numeral System Base symbols:

I = 1; V = 5; X = 10; L = 50; C = 100; D = 500; M = 1000

Notation:

It starts on the left with the symbol of the largest number; the symbols I, X, C are written no more than three times; if the symbol of a smaller number precedes that of a larger number (e.g. IV = 4), its value is subtracted from the larger one, however this is only valid for CM, XC, IX, IV .

Example:

1998 is equivalent to MCMXCV III (MIIM is not valid).

3.2 Elementary Calculus 3.2.1 Elementary Foundations Fundamental arithmetic operations for a, b, c ∈ R

a

b

c

addition

a+b = c

summand

summand

sum

subtraction

a−b = c

minuend

subtrahend

difference

multiplication

a·b = c

factor

factor

product

division

a =c b

dividend, numerator

divisor, denominator

quotient, fraction

25

3.2 Elementary Calculus 3.2.1.1 Axioms Commutative law a + b = b + a

a·b = b·a

Associative law

(a + b) + c = a + (b + c)

(a · b) · c = a · (b · c)

Distributive law

a · (b + c) = a · b + a · c

Sign conventions (+a) · (+b) = (−a) · (−b)

a, b > 0

(+a) · (−b) = (−a) · (+b) (−a) a (+a) = =+ (+b) (−b) b

(+a) (−a) a = =− (−b) (+b) b

3.2.1.2 Factorisation Remark: multiplication and division before addition and subtraction a + (b + c − d) = a + b + c − d ac + bc = c · (a + b)

a − (b + c − d) = a − b − c + d

ac − bc = c · (a − b)

−ac − bc = −c · (a + b)

a · (b − c) = ab − ac

a · (b + c) = ab + ac

(a + b) · (c + d) = ac + ad + bc + bd

(a − b) · (c − d) = ac − ad − bc + bd

(a + b) · (c − d) = ac − ad + bc − bd

(a − b) · (c + d) = ac + ad − bc − bd

3 Arithmetic

26 3.2.1.3 Relations a < b ⇔ b > a ⇔ (b − a) > 0 a < b and c > 0 ⇒ a+c < b+c ⇒ a·c < b·c a < b and a > 0 ⇒ −a > −b ⇒

1 1 > a b

3.2.1.4 Absolute Value, Signum Definitions: absolute value of a (|a|)

signum of a (sgn a)

a>0

|a| = +a

sgn a = 1

a=0

|a| = 0

sgn a = 0

a1

33

3.2 Elementary Calculus n

n

∑ ai · bi ̸= ∑ ai · ∑ bi

i=1

i=1

n

n

n>1

c = constant

∑ cai = c ∑ ai

i=1

i=1

m

n

n

∑ ai + ∑

i=1

ai =

i=m+1

∑ ai ,

m 0 

special:

a2n+1 < 0

(−1)2n

=

1

(−1)2n+1

=

−1

a = constant; a ∈ R+ 0 n; m ∈ Z∗ 1

an = n

1

a m = (am ) n =

√ n

Digression: √ n

a=x

√ 2

a=

√

√ a

a2 = |a|

√ n √ n

0=0 1=1

⇔

xn = a

a = radicand

n = order of the root

√ n

a

am

3 Arithmetic

36 Theorems: with (a, b, p, q ∈ R, m, n ∈ Z) am · an = am+n  a n an = ; b ̸= 0 n b b q q √ n √ m √ m n a= a = m·n a √

n·k

am·k =

√ n

am

an · bn = (a · b)n

am = am−n an

(am )n = am·n

pan ± qan = (p ± q) · an

√ √ √ n n n a· b = a·b

r √ n a a √ = n n b b

√ √
m m n m a = n a = an

Rationalisation of the Denominator If there is an algebraic function (= root with argument) in the denominator of a fraction, it may under certain circumstances make sense to extend the fraction in such a way that the denominator becomes rational. Examples: √ √ √ √ 4·x 4·x 3 x 4·x 3 x 4·x· 3 x √ √ √ √ = 3 · 3 = 3 = = 4· 3 x 3 2 2 3 x x x x x √ √ 2 2 (a − b) 2 · (a − b) √ = √ · √ = √ √ = 3 · (a + b) 3 · (a + b) (a − b) 3 · (a + b) · (a − b) √ 2 · (a − b) = 3 · (a2 − b)

37

3.2 Elementary Calculus

3.2.5 Logarithms Definition: The logarithm of b (numerus) to the base a is the real number c (exponent). loga b = c ⇔ b = ac

a, b ∈ R+ , a ̸= 1

Every equation ax = b has exactly one real solution.

Rules:

loga a = 1

loga (ab ) = b

b∈R

loga 1 = 0

loga x < 0

for x < 1

loga x > 0

for x > 1

3 Arithmetic

38 Examples: 1) 3x = 81

x = log3 81 = 4 test: 34 = 81

2) log5 0.008 = −3

test: 5−3 = 0.008

3) log253 100 = 0.8323

test: 2530.8323 = 100

Logarithmic Laws (a, u, v ∈ R+ , a ̸= 1) 1) loga (u · v)

= loga u + loga v

1 u

= loga u−1

= − loga u

3) loga ur

= r · loga u

r∈R

√ 4) loga r u

 1 = loga u r

=

2) loga

1 loga u r

Logarithmic Systems: Common Logarithm Base a = 10

Notations:

log10 b = lg b lg b = c

⇔ b = 10c

lg 10k = k

k∈R

39

3.2 Elementary Calculus Natural Logarithm Base a = e lim 1 + 1n

e = Euler’s number

Notations:

n→∞


n

= 2.718281828459

loge b = ln b

ln = “logarithmus naturalis”

ln b = c

⇔ b = ec ,

ln ec = c

c∈R

b>0

Logarithm to an Arbitrary Base logk lna = loga lnk

Notations:

loga k =

Example:

log4711 15 =

ln15 log15 = = 0.3202 log4711 ln4711

3.2.6 Factorial n

n ∈ N∗

(read: n factorial)

Recursion formula:

(k + 1)!

= k! · (k + 1)

Definitions:

0! = 1

1! = 1

n! = 1 · 2 · 3 · ... · n =

∏i i=1

k∈N

3 Arithmetic

40

3.2.7 Binomial Coefficient (read “n choose k”)

For n, k ∈ N :

For n, k ∈ N :

      

n!   n = k!(n − k)!  k      0

for 0 ≤ k ≤ n

for 0 ≤ k ≤ n

      0 n n = = =1 0 0 n     n n = =n 1 n−1

Pascal’s Triangle for Determining the Binomial Coefficients n=0 n=2

1

n=3

1

n=4 n=5

row sum

1

1

2 3

22

1 3

23

1

+

+

+

4

6

4

20

24

1

1

5

10

10

5

1

  5 0

  5 1

  5 2

  5 3

  5 4

  5 5

25

41

3.3 Sequences

The boundary values are always 1, the mean values are the sum of the values immediately above them (left and right). Examples:   7 7! 1·2·3·4·5·6·7 = = = 21 5 5! · (7 − 5)! (1 · 2 · 3 · 4 · 5) · (1 · 2) 

− 13 2





− 13 · − 13 − 1 (− 13 ) · (− 43 ) 2 = = = 0.2 = 2! 1·2 9

3.3 Sequences 3.3.1 Definition A sequence ak is a mapping of natural numbers, k ∈ N∗ (possibly also k ∈ N) to a universal set SU , ak ∈ R: ak = a1 , a2 , a3 , ..., ak

k ∈ N∗ ; ak ∈ R

If S corresponds to a set of points, a so-called point sequence is created; if S corresponds to a set of numbers, a so-called numeric sequence is created.

A real numeric sequence is an ordered set of real numbers. It corresponds to a discrete function of the mapping: ak = f (k)

with D f = N∗ and R f = R

3 Arithmetic

42 Sequences can be finite or infinite.

Finite sequences have a last term an : with ai = 0 for all i > n

ak = a1 , ..., a3

Infinite sequences have an unlimited number of terms: ak = a1 , a2 , ...

Examples: (1)

ak = k 3 ⇒ ak = 1, 8, 27, 64, 125, ... a5 = 5 th term = 125

(2)

ak = (−1)k · (ak + 1) ⇒ ak = −2, 3, −4, 5, −6, ... a5 = −6

(3)

Sequences with alternating signs: (3a) ak = (−1)k+1 = +1, −1, +1, −1, ... (3b) ak = (−1)k = −1, +1, −1, +1, ...

Fundamental Terms: A numeric sequence ak is called negatively definite

ak < 0

43

3.3 Sequences monotonically increasing

ak ≤ ak + 1

strictly monotonically increasing

ak < ak + 1

monotonically decreasing

ak ≥ ak + 1

strictly monotonically decreasing

ak > ak + 1

bounded above (Bu = upper bound)

ak ≤ Bu ; Bu ∈ R

bounded below (Bl = lower bound)

ak ≥ Bl ; Bl ∈ R

bounded

Bl ≤ ak ≤ Bu

constant

ak = ak+1

Supremum, Infimum, Limits The supremum of an upwards bounded sequence ak , sup ak , is the least upper bound (= the upper limit) of ak . Example: ak = −k3 ⇒ ak = −1, −8, −27, −64, −125, ...

Possible upper bounds are e.g. 17 or 0 or −1. However, the supremum (= the upper limit) of ak is definitely: sup ak = −1

The infimum of a downwards bounded sequence ak , inf ak , is the greatest lower bound (= the lower limit) of ak .

3 Arithmetic

44 Example: ak = k3 ⇒ ak = 1, 8, 27, 64, 125, ...

Possible lower bounds are e.g. −100 or 0 or 1. However, as infimum (= the lower limit) only one definite value exists: inf ak = 1

3.3.2 Limit of a Sequence The sequence ak is called convergent with the limit g, if for any real, positive number ε nearly all sequence terms ak lie within the ε-range of g, ]g − ε; g + ε[: |ak − g| < ε

for nearly all k ∈ N∗ ;

lim ak = g or

ak k→∞

k→∞

ε ∈ R+

→g

Read: The limit of ak for k towards infinity is equal to g. If a numerical sequence ak has the limit g, ak is called convergent, ak converges towards g. If no limit exists, ak is divergent. Theorems: For lim ak = g1 and lim bk = g2 applies: k→∞

k→∞

(1) lim (ak ± bk ) = g1 ± g2 k→∞

(2) lim (ak · bk ) = g1 · g2 k→∞

(3) If, apart from the starting terms, all bk ̸= 0 and g2 ̸= 0,

45

3.3 Sequences the following applies: (4) lim (ank ) = gn1 k→∞

ak g1 = k→∞ bk g2 lim

n ∈ N∗

(5) Every convergent sequence is limited. Remark: Not every limited sequence is convergent. The limited sequence −1, +1, −1, +1, ... for example is divergent. (6) Every limited and monotone sequence is convergent. (7) ak ≤ bk ⇒ g1 ≤ g2

Null Sequence A sequence ak is called null sequence, if its limit is zero: lim ak = 0

k→∞

Example: ak =

1 is a null sequence for k ∈ N∗ , in that case: lim ak = 0 k→∞ k

Improper Limit ak diverges towards ∞ or −∞: lim ak = ∞

k→∞

lim ak = −∞

k→−∞

3 Arithmetic

46

Examples of Limits of Selected Numerical Sequences (k ∈ N∗ )

lim ak = 0

k→∞

  1 k lim 1 + = e = 2.718281828459... (number) k→∞ k

  1 1 1 lim 1 + + + ... + − ln k = C = 0.57721 k→∞ 2 3 k k!

lim

k→∞ kk · e−k ·

lim

k→∞

√ k

√ = k

a=1

(Euler’s constant)

√ 2π

(Stirling’s formula)

a>0

lim

k→∞

√ k

k=1

3.3.3 Arithmetic and Geometric Sequences Arithmetic Sequences In an arithmetic sequence, the difference d between each two consecutive terms of a sequence ak is constant: ak+1 − ak = d

with

d = constant for all k ∈ c,

the numerical sequence is arithmetical.

Geometric Sequence In a geometric sequence, the quotient q between each two consecutive terms of a sequence ak is constant:

3.4 Series ak+1 =q ak

47 with

q = constant for all k ∈ N∗ ,

the numerical sequence is geometrical.

3.4 Series 3.4.1 Definition A series sn (to a sequence ak ) is equivalent to the n th partial sum of the first n terms (summands) of the sequence ak : n

sn = a1 + a2 + ... + an =

∑ ak k=1

3.4.2 Arithmetic and Geometric Series Arithmetic Series n

sn = a1 + a2 + a3 + ... + an =

∑ ak

with

d = an − an−1 = ...

k=1

= a3 − a2 = a2 − a1 = constant k th term:

ak = a1 + (k − 1) · d

last term:

an = a1 + (n − 1) · d

sum:

sn =

n n (a1 + an ) = [2a1 + (n − 1) · d] 2 2

3 Arithmetic

48 Arithmetic Series of Higher Order

An arithmetic series of the i th order is present if only the i th difference sequence has constant terms: ak = bi (k − 1)i + bi−1 (k − 1)i−1 + ... + b0

with: k = 1, ...

Example: ak

=

1

△1 ak =

5

10

4

5

8

13

1

3

5

△2 ak = △3 ak =

2

18

2

31

7 2

51

basic sequence

...

20 ...

1st difference sequence

...

2nd difference sequence 3rd difference sequence

...

The primary sequence ak is equivalent to an arithmetic series of the 3rd order.

Geometric Series A geometric series of the i th order is present if the quotient q of two consecutive terms is constant. n

∑ q0 qk =

For |q| < 1:

n=0

with

q=

k=0

=

q0 1−q

an an−1 a3 a2 = = ... = = = constant an−1 an−2 a2 a1

3.4 Series

49

Example: ao = 5; a1 = 15, a2 = 45, a3 = 135, . . . n

⇒

∑

5 · 3k = 5 · 30 + 5 · 31 + 5 · 32 + 5 · 33 + . . .

k=0

= 5 + 15 + 45 + 135 + . . . = ∞ with

q=3

because

15 45 = 3; = 3 etc. 5 15

Infinite Geometric Series s = 1 + x + x2 + x3 + . . . + xn

s=

1 − xn+1 1−x

with

x ̸= 1

Example:

etc. 1 2

1

s=


1 n+1 2 1 − 12

1−

=

1 1 − 2n+1

0.5

if n → ∞, the following is valid:

1 4

1 8

3 Arithmetic

50 lim s =

n→∞

1 = 2, 0.5

i.e. in this example, the number 2 will never be reached.

Chapter 4

Algebra 4.1 Fundamental Terms Variables are placeholders (e.g. a, b, x, y, ...) that can be replaced by numbers from a given universal set SU . A term within a universal set SU is an expression composed of variables, numbers and/or arithmetic symbols. Division by zero is not possible. Examples: (1)

6

(2)

8−2

(3)

x+2

(4)

3b + 5

(5)

x2 − 4x + 6

with

a, b, x, y, ∈ R

Equations and Inequations An equation (inequation) is created when terms are connected by the equal sign “=” (the not-equal signs “, ≥” or “̸=”). The solution set SS of an equation/inequation is the set of elements that makes the initial form a true statement when used in place of the variables. The equation/inequation has no solution if the solution set is equal to the empty set.

© Springer-Verlag GmbH Germany, part of Springer Nature 2023 F. W. Peren, Math for Business and Economics, https://doi.org/10.1007/978-3-662-66975-4_4

51

4 Algebra

52 Examples: (1)

(x − 5)(x − 3) = 0

x∈R

SS = {5; 3}

(2)

x+1 ≤ x

x∈R

SS = 0/

Universal Equations If a solution set SS is identical to the universal set SU , the equation is generally valid with respect to the universal set SU . Examples: (1) (2)

x∈R

2(x + 1) = 2x + 2 2

2

(a + b) = a + 2ab + b

2

a, b ∈ R

Equivalent Transformations of Equations Equations are equivalent if their solution sets are identical. With nonequivalent transformations (square/multiply/divide by terms containing the variable(s)) other solution sets can arise. Sample offered!

x∈R

Examples:

5 · (x − 1)

= 30

⇔

x−1

=6

(equivalent transformation)

⇔

x

=7

SS = 7

x−2

=

(1)

(2) ⇒ ⇔

√

2

=x

2

=0

x − 4x + 4 x − 5x + 4

x (no equivalent transformation)

53

4.2 Linear Equations   s 2 5 5 = ± − 4 (p/q formula) 2 2

⇔

x1,2

⇔

x1 = 2.5 +

⇔

x2 = 2.5 −

q

(2.5)2 − 4 = 4

q

(2.5)2 − 4 = 1

Sample offered!

⇔

4 is the only solution of equation (2).

⇔

SS = {4}

4.2 Linear Equations 4.2.1 Linear Equations with One Variable Normal form:

ax + b = 0 ;

x∈R

a ̸= 0

Equivalent transformations are used to separate the correct variable. Example: 50x + 40 ⇔ 60x ⇔ x

= −10x = −40 2 = − 3

 SS =

2 − 3



Fractional Equations The domain of definition corresponds to the base set excluding the values where the denominator becomes zero.

4 Algebra

54 Example: 5 x−3

=

2 x−1

⇔ 5 · (x − 1)

= 2 · (x − 3)

⇔ x

1 = − 3

D = R\{1; 3}

 SS =

1 − 3



Fractional Inequations with One Variable For the transformation of fractional inequations, a case distinction must take place which is provided by the domain of definition. Definition gaps can accordingly be divided into cases that are to be investigated separately.

Example:

x−2 −3

Case 1

positive denominator

Case 2

negative denominator x < −3

55

4.2 Linear Equations Step 3:

Solving inequations separately Case 1:

x > −3

x−2 x+3

x

| −x; −6

Define intersections 1. Fall: x > −3 ∩ x > −8

Intersection:

x > −3

2. Fall: x < −3 ∩ x < −8

Intersection:

x < −8

4 Algebra

56 Step 5:

Define union of sets/solution set x > −3

∪

x < −8

SS = {x | x < −8 ∪ x > −3}

4.2.2 Linear Inequations with One Variable When multiplying with or dividing by a negative number, the relation sign is reversed. Example:

−3x − 10 < 2 · (x + 20) ⇔ −3x − 10 < 2x + 40

| −2x; +10

⇔ −5x

< 50

| ÷(−5)

⇔ x

> −10

| inversion of the relation sign

⇔ SS = {x x > −10} = ] − 10; ∞[

4.2.3 Linear Equations with Multiple Variables A definite determination of n variables is only possible if n independent equations exist (unambiguously determinable equation system). If there are only r independent equations with n variables (r < n), then (n − r) variables exist as free parameters and thus an infinite number of number tuples as solutions.

57

4.2 Linear Equations Examples: (1)

3x + 8y = 100

⇔

x=

x, y ∈ R

100 − 3x 100 − 8y resp. y = 3 8

A definite solution is not possible.

(2)

(a)

3x + 8y =

100

(b)

x + 2y =

50

| −2y

⇔

x =

50 − 2y

| (b) resolved to x

(b) in (a) 3 · (50 − 2y) +8y = | {z }

100

x

⇔

y =

−25

y in (b)

x =

50 − 2 · (−25) | {z } y

⇔

x =

100

There is a definite solution, namely:

SS = {(x, y) | x = 100; y = −25}

4.2.4 Systems of Linear Equations A system of linear equations consists of several linear equations. Its solution set is the set of all (ordered) tuples of values for which all equations become true statements.

4 Algebra

58

Equivalent Transformations of Systems of Linear Equations (1)

Multiplication of an equation by a real number,

(2)

Addition of the multiple of one equation to another equation (linear combination),

(3)

Swapping equations.

Solving Systems of Linear Equations (a) Substitution Method An equation is solved for a variable. This term is inserted into another equation in the place of the corresponding variable. Example: (a)

x + 2y

= 15

(b)

2x − 2y

= 24

| +2y; ÷2

⇔

x

= 12 + y

| (b) resolved to x

12 + y +2y | {z }

= 15

| −12

⇔

3y

=3

| ÷3

⇔

y

=1

(b) in (a)

x

replacing the y value in (b), results in the x value:

y in (b)

x

= 12 + |{z} 1 y

59

4.2 Linear Equations ⇔

= 13

x

SS = {(x, y) | x = 13; y = 1}

(b) Equalisation Method Two equations are solved for the same variable and the terms on the right sides are set equal. Example: (a)

y

= −3x + 900

(b)

y

= x + 200

equalling (a) to (b):

x + 200

= −3x + 900

| +3x; −200

⇔

4x

= 700

| ÷4

⇔

x

= 175

(a) = (b)

replacing the x value in (a) or (b) results in the y value:

x in (b)

y

= |{z} 175 +200

⇔ y = 375

= −3 · |{z} 175 +900

⇔ y = 375

x

or

x in (a)

y

x

SS = {(x, y) | x = 175; y = 375}

4 Algebra

60 (c) Addition Method

Two equations are multiplied with or divided by suitable real numbers in such a way that one variable is eliminated by adding or subtracting the two equations. Example:

(a)

3x

+

2y

=

15

(b)

x

−

y

=

12

(a)

3x

+

2y

=

15

(b)

−3x

+

3y

=

− 36

(a) + (b)

0x

+

5y

=

− 21

=

4.2

y

| ·(−3)

| ÷5

Replacing the y value in (a) or (b) results in the x value: 15 − 2 · (−4.2) = 7.8 3

y in (a)

x=

y in (b)

x = 12 + (−4.2)

= 7.8

SS = {(x, y) | x = 7.8; y = −4.2}

61

4.2 Linear Equations

4.2.5 Linear Inequations with Multiple Variables The solution set is the average (of the intersection) of the solution sets of the individual inequations. Example: x + 2y − 4 ⇔

y

< 0

∧

y ≥ −1.5

1 < 2− x 2

∧

y ≥ −1.5

x, y ∈ R

The solution set includes all points of the Cartesian coordinate system bounded by the straight lines y = 2 − 0.5x ∧ y = −1.5. If the (boundary) line itself is excluded (relation signs , ̸=), it has to be drawn dashed (as in the example above).

4 Algebra

62

4.3 Non-linear Equations 4.3.1 Quadratic Equations with One Variable a − b − c Formula General form: a2 + bx + c = 0

x1,2

√ −b ± b2 − 4ac = 2a

=

x1,2

a, b, c ∈ R a, b ̸= 0 q
b 2 − 2ac − b2 ± 2 = a

p/q Formula Normal form: x2 + px + q = 0 (right side is equal to 0 absolute term of x2 is equal to 1) p x1,2 = − ± 2

r  p 2 2

−q

(p/q formula)

SS = {x1 ; x2 } If the radicand is negative, the following applies: SS = { } Example: 2x2 − 8x = −6 ⇒ x2 −4 x +3 = 0 |{z} |{z} q

p

⇒ p = −4;

 s  −4 −4 2 =− ± −3 2 2 

x1,2

q=3

(initial equation) (normal form)

63

4.3 Non-linear Equations √ x1,2 = 2 ± 4 − 3 ⇒ x1 = 2 + 1 = 3 x2 = 2 − 1 = 1

SS = {3; −1}

Completing the Square Normal form:

x2 + px + q = 0

(right side is equal to 0 absolute term of x2 is equal to 1)

x2 + px = −q Both sides are completed “quadratically”, i.e. with a summand which arises from the first binomial form:

(a + b)2 = a2 + 2ab + b2

b2 is added, whereby b is obtained from the second summand: px = ˆ 2ab ⇒ ⇔

x= ˆ a∧p= ˆ 2b p b= 2

x2 + |{z} a2

px + |{z} 2ab

 p 2

= −q + | 2{z } b2

Complementary term

⇔

⇒

  p 2 p 2 x+ = −q + 2 2 r  p p 2 x1,2 + = ± −q 2 2

(first binomial form)

 p 2 2

4 Algebra

64

⇔

p =− ± 2

x1,2

r  p 2 −q 2

(corresponds to p/q formula)

Example: 15 x = 10 2

⇔

5x2 −

⇔

3 x2 − x − 2 = 0 2

⇔

x2 −

(initial equation) (normal form)

3 x =2 2 |{z} 2ab

   2 3  2  2    = 3  2  4 ⇒

x2 −

 2  2 3 3 3 x+ = 2+ 2 4 4

 ⇔

⇔

⇒

x−

3 4

2 = 2+

3 x1,2 − = ± 4

x1,2

3 = ± 4

=

(complementary term)

 2 3 4

(second binomial form)

s  3 2 +2 2

s  3 2 +2 2

3 √ ± 2.5625 4

(corresponds to p/q formula)

65

4.3 Non-linear Equations ⇒

x1 =

3 √ + 2.5625 = 2.3508 4

x2 =

3 √ − 2.5625 = −0.8508 4

SS = {−0.8508; 2.3508}

4.3.2 Cubic Equations with One Variable General form:

a3 x3 + a2 x2 + a1 x + a0 = 0

ai ∈ R

Normal form:

x3 + ax2 + bx + c = 0

a, b, c ∈ R

Solving Cubic Equations with One Variable (1)

The first x that leads to the solution of the normal form is obtained by trial and error. This can be de facto simplified by choosing an integer divisor of the absolute term c as divisor.

(2)

Polynomial long division ⇒ quadratic equation

(3)

p/q formula / completing the square

Example:

(1)

y = x3 − 3x2 − 25x − 21

x1 = −3, since (−3)3 − 3 · (−3)2 − 25 · (−3) − 21 = 0

4 Algebra

66 (2)

(x3 −3x2 −25x −21) ÷(x + 3) =

x2 − 6x − 7 | {z }

quadratic equation 3

2

− (x +3x ) −6x2 −25x − (−6x2 −18x) −7x

−21

− (−7x −21) 0  s  6 6 2 =− − ± − +7 2 2 √ = 3 ± 16 

(3)

x2/3 x2/3

x2 = 3 + 4 = 7 x3 = 3 − 4 = −1 SS = {−3; −1; 7}

Solving Cubic Equations with One Variable without Absolute Term (1)

Factorise x to the smallest degree ⇒ First solution: x = 0 and quadratic equation

(2)

p/q formula / completing the square

Example: (1)

x8 + 2x7 − 8x6 = 0 x6 (x2 + 2x − 8) = 0

x6 is equal to zero when x is zero:

⇒ x1 = 0

67

4.3 Non-linear Equations (2)

x2 + 2x − 8 = 0

x2/3

2 =− ± 2

s  2 2 +8 2

√ x2/3 = −1 ± 9 x2 = −1 + 3 = 2 x3 = −1 − 3 = −4

4.3.3 Biquadratic Equations General form:

a4 x4 + a2 x2 + a0 = 0

ai ∈ R

Normal form:

x4 + ax2 + c = 0

a, c ∈ R

Solving Biquadratic Equations without Absolute Term (1)

Substitution z2 + cz + d = 0

z = x2 (quadratic equation)

(2)

p/q formula / completing the square

(3)

Solve for z

(4)

Resubstitution (z → x2 )

(5)

Solve for x

4 Algebra

68 Example: (1) (2)

x4 − x2 − 6 = 0

denote x2 = z

z2 − z − 6 = 0 s  1 1 2 z1/2 = ± − +6 2 2 √ z1/2 = 0.5 ± 6.25

p/q formula

z1 = 0.5 + 2.5 = 3 denote z = x2

z2 = 0.5 − 2.5 = −2 (3)

x2 = −2 ∨ x2 = 3

√ −2 ∨ x = − −2 √ √ ∨ x= 3 ∨ x=− 3

⇔ x=

√

∨

√ √ i.e. (since x = −2 ∨ x = − −2 is not defined): √ √ SS = {− 3; 3}

4.3.4 Equations of the nth Degree General form of an algebraic equation of the n th degree an xn + an−1 xn−1 + ... + a1 x + a0 = 0

a1 ∈ R, an ̸= 0

There are no general solutions for general equations of 5 th and higher degree. n th degree polynomials: An n th degree algebraic equation becomes an n th degree polynomial (= n th degree polynomial function), when: an xn + an−1 xn−1 + ... + a1 x + a0 = 0

a1 ∈ R, an ̸= 0, n ≥ 5

69

4.3 Non-linear Equations

4.3.5 Radical Equations The variable x appears within the radicand (the term from which the square root is extracted). To eliminate roots, non-equivalent transformations (= exponentiation) are necessary. This results in equations of which solutions do not necessarily have to be solutions of the original equation. Sample offered! Basic equation √ n

√ √

x = an

ai ∈ R x ∈ R, whereby the whole radicand with even n must not be negative. x = variable

x+b = a ⇒

x = a2 − b

x ≥ −b, a ≥ 0

cx + b = a ⇒

x=

⇒

x=a

(a − b)2 c

sgn x = sgn c ; c ̸= 0

The necessary condition for the domain is that all radicands ≥ 0.

Examples: (1) ⇔

√ √

2x − 3 − 5 = 0

| +5

2x − 3 = 5

| ()2 ; +3; ÷2

⇒ x = 14 Test:

√

2 · 14 − 3 − 5 = 0

SS = {14}

4 Algebra

70 (2)

√

√ √ x − 1 + x + 6 = 5x − 1

√ √ √ ( x − 1 + x + 6)2 = ( 5x − 1)2 √ √ (x − 1) + 2 x − 1 x + 6 + (x + 6) = 5x − 1 | {z }

| −(x − 1); −(x + 6)

binomial form

√ √ 2 x − 1 x + 6 = (5x − 1) − (x − 1) − (x + 6) | ÷2 √

√ 3x − 6 x−1 x+6 = = 1.5x − 3 2

| ()2

(x − 1)(x + 6) = (1.5x − 3)2

| binomial form

x2 + 6x − x − 6 = 2.25x2 − 9x + 9

| −2.25x2 ; +9x; −9

−1.25x2 + 14x − 15 = 0

| ÷(−1.25)

x2 − 11.2x + 12 = 0 s  11.2 11.2 2 ± − 12 x1 = − 2 2 2 √ x 1 = 5.6 ± 19.36

| p/q formula

2

x1 = 10; x2 = 1.2

Test:

√ √ √ 10 − 1 + 10 + 6 − 5 · 10 − 1 = 0 √ √ √ 1.2 − 1 + 1.2 + 6 − 5 · 1.2 − 1 ≈ 0.894 ̸= 0

SS = {10}

71

4.4 Transcendental Equations

4.4 Transcendental Equations Every non-algebraic equation is called transcendental.

4.4.1 Exponential Equations The variable appears in the exponent.

ax = b

Basic equation

a, b ∈ R+ ; a ̸= 1 x∈R x = variable

(1)

(2)

Solution:

log ax = log b

⇔

x · log a = log b

⇔

x=

log b lg b ln b = = log a lg a ln a

If the bases are the same, then: ax = ac

⇒ x=c

The choice of the base plays no role.

4 Algebra

72 Examples: (1)

5x+1 = 18

⇔

log(5x+1 ) = log 18

⇔

(x + 1) · log 5 = log 18

⇔

(x + 1) =

⇔

log 18 log 5 log 18 − 1 ≈ 0.7959 x= log 5 SS = {0.7959}

(2)

√ 3

ax−1 =

x−1 3

√

ax+3

x+3 2

⇔

a

⇔

x−1 x+3 = 3 2

⇔

2(x − 1) = 3(x + 3)

⇔

2x − 2 = 3x + 9

⇔

x = −11

=a

SS = {−11}

The choice of the base plays no role.

73

4.4 Transcendental Equations

4.4.2 Logarithmic Equations The argument is presented in logarithmic form. Basic equation

Solution:

loga x = b a, x ∈ R+ x = variable (the argument)

x = ab

The solution is not equivalent regarding the (original) domain of definition. If the base of the logarithm is the same, then: ⇒x=c

loga x = loga c

Examples: (1)

ln(2x − 5) = 25 Domain of definition  2x − 5 > 0

⇒

D=

x|x>

ln(2x − 5) = 25 ⇔

eln(2x−5) = e25

⇔

2x − 5 = e25

⇔

x=

1 25 (e + 5) ≈ 3.6 · 1010 2

SS = {3.6 · 1010 }

5 2



| extend with e

4 Algebra

74 (2)

log(x2 + 1) = 2 log(x + 2) Domain of definition x2 + 1 > 0

⇒

−∞ < x < ∞

∧

x+2>0

⇒

x > −2

⇒

D = {x | −2 < x < ∞} ln(x2 + 1) = 2 ln(x + 2)

⇔

ln(x2 + 1) = ln(x + 2)2

⇔

eln(x

⇔

x2 + 1 = (x + 2)2

⇔

x2 + 1 = x2 + 4x + 4

⇔

−3 = 4x   3 SS = − 4

2 +1)

= eln(x+2)

| extend with e

2

Remark: The transformation of a logarithmic equation can lead to the fact that the domain of definition is no longer equivalent. Example:

⇒

x=x

with x ∈ R

ln x = ln x

with x ∈ R+

The domain of definition has changed during the transformation, thus no equivalence is given.

4.5 Approximation Methods

75

4.5 Approximation Methods The following iteration methods are used to determine a zero of nonlinear equations. Solution principle: The zero x0 of a (in the relevant interval) continuous real function f = f (x) is derived from the solution of the equation f (x) = 0.

4.5.1 Regula falsi (Secant Method) Condition: f = f (x) is a (in the relevant interval) continuous, real function with a single zero x0 Principle: The zero X0 is between two (start) values Xb and Xa with f (xb ) · f (xa ) < 0

Fig. 4.1: Regula falsi (Secant Method)

76

4 Algebra

The (non-linear) curve is geometrically replaced by the secant between Pb and Pa (secant method, Fig. 4.1). The point xi , i.e. the intersection of the secant with the x-axis, is calculated as follows:

xi = xb − f (xb ) ·

xa − xb f (xa ) − f (xb )

By the (iterative) repetition, the secant gradually gets closer to the zero x0 , so that finally x0 can be determined (approximately). The regula falsi method is numerically stable, i.e. the error decreases or remains the same from iteration to iteration.

Example: f (x) = x3 + 2x − 1

zero: x0 = ?

First iteration: Arbitrary choice of two start values with f (xb1 ) · f (xa1 ) < 0; i.e. one start value is to the left, one to the right of the zero x0 that is searched for: xb1 = 0

f (xb1 ) = −1

xa1 = 1

f (xa1 ) = 2

⇒ xi1 = 0 + 1 ·

1−0 1 = ≈ 0.333 2+1 3

The iterative repetition of the secant procedure leads to further (successive) approach to x0 :

4.5 Approximation Methods

77

Second iteration: around the value x = 0.333, e.g.:

xb2 = 0.2

f (xb2 ) ≈ −0.592

xa2 = 0.5

f (xa2 ) ≈ 0.125

⇒ xi2 = 0.2 + 0.592 ·

0.5 − 0.2 ≈ 0.4477 0.125 + 0.592

Third iteration: around the value x = 0.4477, e.g.:

xb3 = 0.43

f (xb3 ) ≈ −0.0605

xa3 = 0.46

f (xa3 ) ≈ 0.0173

⇒ xi3 = 0.43 + 0.0605 ·

0.46 − 0.43 ≈ 0.4533 0.0173 + 0.0605

The zero x0 can be defined more precisely after each iteration. In this example, it is approximately: x0 ≈ 0.4534; f (x0 ) ≈ 0.0000061453.

4.5.2 Newton’s Method (Tangent Method) Condition: The function f = f (x) is differentiable twice in the interval [x1 ; x2 ]. Within this interval, it has a zero x0 with f ′ (x0 ) ̸= 0. There is an environment V around x0 so that the Newton’s method is applicable and converges for each start value xi ∈ V toward x0 .

78

4 Algebra

The procedure fails if the curve of f = f (x) at the respective approximation point is (almost) parallel to the x-axis. Iteration rule

xi+1 = xi −

f (xi ) f ′ (xi )

f ′ (x0 ) ̸= 0

necessary condition,

| f (x) · f ′′ (x) | < ( f ′ (x))2

sufficient condition.

The (non-linear) curve of the function f = f (x) is geometrically replaced by its tangent at the respective point Pi (tangent method, Fig. 4.2). The starting value x1 and thus also P1 can be arbitrarily selected within the interval [x1 ; x2 ].

Fig. 4.2: Newton’s Method (Tangent Method)

79

4.5 Approximation Methods

The respective xi+1 value is determined by the (aforementioned) respective tangent of f at point Pi with the intersection of the abscissa. xi+1 then forms the start value of the subsequent iteration. By the (iterative) repetition, the tangent gradually gets closer to the zero x0 , so that finally x0 can be determined (approximately). The Newton’s method is numerically stable. Order of convergence ρ = 2 Example: f (x) = x3 + 2x − 1 Zero x0 = ? f ′ (x) = 3x2 + 2

f ′′ (x1 ) = 6x

First iteration: Arbitrary selection of a start value: x1 = 1 ⇒

f ′ (x1 ) = 5;

f (x1 ) = 2;

f ′′ (x1 ) = 6x

sufficient condition: | f (x1 ) · f ′′ (x) | < ( f ′ (x1 ))2 =

| 2 · 6 | < 25

⇒

x2 = x1 −

f (x1 ) 2 3 = 1− = ′ f (x1 ) 5 5

o.k.

80

4 Algebra

The iterative repetition of the secant procedure leads to further (successive) approach to x0 : Second iteration: x2 = ⇒

3 5

f (x2 ) = 0.416;

f ′ (x2 ) = 3.08;

f ′′ (x2 ) = 3.6

sufficient condition: | f (x2 ) · f ′′ (x2 ) | < ( f ′ (x2 ))2 =

| 0.416 · 3.6 | < 9.4864

⇒

x3 = x2 −

o.k.

f (x2 ) 3 = − 0.135 = 0.469 f ′ (x2 ) 5

etc. The zero x0 can be defined more precisely after each iteration. In this example, it is approximately: x0 ≈ 0.4534; f (x0 ) ≈ 0.0000061453.

4.5.3 General Approximation Method (Fixed-point Iteration) Condition: f = f (x) is a (in the relevant interval) continuous, real function with a single zero x0 . The general approximation method (fixed-point iteration) is shown graphically in Fig. 4.3. Principle: f (x) = 0 is transformed to x = g(x) (fixed-point equation), whereby g(x) is a real, continuous and differentiable function (in the relevant interval).

4.5 Approximation Methods

81

Iteration rule xi+1 = g(xi )

with | g′ (x) |< 1

If 0 ≤ g′ (x) < 1, the convergence is monotonous, i.e. one approaches the searched zero x0 permanently from the same side.

Fig. 4.3: General Approximation Method (Fixed-point Iteration) If −1 < g′ (x) < 0, the convergence is oscillating, i.e. two successive approximate values lie on different sides of the zero x0 . The procedure fails if | g′ (x) |> 1 because the inclination angle of the tangent with respect to the curve of g is not between 0◦ to 45◦ resp. 135◦ to 180◦ . As a result, the (“approximation”) values will successively deviate from x0 . The procedure diverges. In this case, f (x) = 0 must be resolved to another term with x (see Example 2).

4 Algebra

82 Example 1: f (x) = x3 + 2x − 1 Zero x0 = ? ⇔

x=

1 − x3 = g(x) 2

g′ (x) = −

3x2 2

First iteration: arbitrary selection of a start value: x1 = 1 g′ (x0 ) =

3 · 12 3 =− 2 2

| g′ (x0 ) |=

3 >1 2

⇒ Condition violated; A new start value required. New first iteration: arbitrary selection of a start value: x1 = 0.5 g′ (x0 ) =

3 · 0.52 = −0.375 2

| g′ (x0 ) |= 0.375 < 1

o.k.; monotonically convergent

4.5 Approximation Methods

83

Second iteration: x2 = g(x1 ) =

1 − 0.53 ≈ 0.4375 2

Third iteration: x3 = g(x2 ) =

1 − 0.43753 ≈ 0.4581 2

Fourth iteration: x4 = g(x3 ) =

1 − 0.45813 ≈ 0.4519 2

etc.

Remark After the third iteration, sufficient accuracy is already reached at the second decimal place.

4 Algebra

84 Example 2: f (x) = x3 + 2x − 8 Zero x0 = ? ⇔

x=

8 − x3 = g(x) 2

g′ (x) = −

3x2 2

First iteration: arbitrary selection of a start value: x1 = 1.5 g′ (x0 ) = −

3 · 1.52 = −3.375 2

| g′ (x0 ) |= 3.375 > 1 divergent ⇒ Resolve f (x) = 0 with the second term of x: x3 = 8 − 2x

⇔

x=

√ 3 8 − 2x = h(x)

2 h′ (x) = − p 3 3 (8 − 2x)2 2 h′ (x0 ) = − p = 0.228 3 3 (8 − 2 · 1.5)2 | h′ (x0 ) |= 0.228 < 1

o.k.; monotonically convergent

4.5 Approximation Methods Second iteration: x2 = h(x1 ) =

√ 3

8 − 2 · 1.5 ≈ 1.710

Third iteration: x3 = h(x2 ) =

√ 3

8 − 2 · 1.710 ≈ 1.661

Fourth iteration: x4 = h(x3 ) =

√ 3

8 − 2 · 1.661 ≈ 1.673

Fifth iteration: x5 = h(x4 ) =

√ 3

8 − 2 · 1.673 ≈ 1.670

etc. x converges to 1.670 and is therefore fixed point = zero of f (x) = x3 + 2x − 1

85

Chapter 5

Linear Algebra Linear algebra is used, among other things, in the analysis of complex business and economic systems.

5.1 Fundamental Terms 5.1.1 Matrix An m × n -matrix A is a rectengular number scheme of m rows and n columns:



a11 a12 . . . a1 j . . . a1n

 a  21   ..  . A=  ai1   .  .  . am1



  a2n   ...     ai2 . . . ai j . . . ain  ⇐ i th row . ..  . .  .. ..  am2 . . . am j . . . amn a22 . . . . ..

a2 j . . . ...

⇑ j th column i = row j = column The ai j ∈ R are called elements of the matrix A. The first index i (i = 1, ..., m) describes the consecutive number of the row, the second index j ( j = 1, ..., n) the consecutive number of the column.

© Springer-Verlag GmbH Germany, part of Springer Nature 2023 F. W. Peren, Math for Business and Economics, https://doi.org/10.1007/978-3-662-66975-4_5

87

5 Linear Algebra

88 Example:

Production conditions in beer production per 100 gallons beer: Output

Lager

Pale Ale

water [gal]

120

140

hops [lbs]

4

6

malt [lbs]

8

3

Input



120 140

 production matrix A =   4 8



 6   3 × 2 -matrix 3

Element: The elements ai j (a21 = 4) indicate how many units of the factor i (i = 1, 2, 3) are needed to produce one unit of the output j ( j = 1, 2) (production coefficient).

5.1.2 Equality/Inequality of Matrices Two matrices Am×n and Bm×n of the same order are called equal if all elements of A and B coincide. A = B if ai j = bi j for all i, j A

< < B if ai j bi j for all i, j > >

A

≤ ≤ B if ai j bi j for all i, j ≥ ≥

5.1 Fundamental Terms

89

Example:

A=

5 7

! B=

7 10

6 7

! C=

9 10

⇒ (A ≤ B)

(B ̸= C)

(A ̸= C)

(B ̸= D)

460 890

! D=

5 71

!

9 10 8

(C < D)

(A ̸= D)

5.1.3 Transposed Matrix If the rows and columns of an m × n-matrix A are swapped, the so-called transposed matrix A′ or At of A with the order n × m is formed.

Example:  A2×3 =

246 135

! ⇒

21



   A′3×2 =  4 3 65

Remark: If a matrix is transposed twice, the original matrix is restored. The following applies: (A′ )′ = A.

5.1.4 Vector An m × 1-matrix is called column vector, a 1 × n-matrix is called row vector. The elements of the vector are called components.

5 Linear Algebra

90 row vector (1 × n):

x = (x1 x2 · · · xn )1×n

column vector (n × 1):   x  1 x   2 x′ =  .   ..    xn n×1

Geometrically, every point P of a k-dimensional space Sk can be described by its k coordinates x1 , x2 , · · · , xk , which can be summarized as a vector. Examples: (1) two-dimensional space

x=

x1

!

x2

=

4

!

3

(2) three-dimensional space   1    x = 3  4

5.1 Fundamental Terms

91

Remarks: • Each matrix Am×n consists of m row vectors and n column vectors.

Example:

159

A=

!

274 row vectors: (1

9); (2

5

column vectors:

1 2

! ;

5

7

! ;

7

4)

9

!

4

• Each 1 × 1-matrix is called scalar. Scalars are considered to be real numbers. Example: The absolute value of a vector [4] = 4. 

x1



  x   2 x= .   ..    xn

or

x′ = (x1

x2

···

xn )

The absolute value of a vector is defined as

q

x12 + x22 + · · · + xn2 .

5 Linear Algebra

92

5.1.5 Special Matrices and Vectors (1) Zero Matrix → all elements of the matrix = zero

 =

0



0

0

0

0

···

0

 0  .  ..  0

0 .. .

0 .. .

0 .. .

···

0

0

0

···

 0  ..  .  0

(2) Zero Vector → all elements of the vector = zero (3) Square Matrices Number of rows = b number of columns ⇒ An×n (4) Main Diagonal of a Matrix The elements a11 , a22 , · · · , ann form the main diagonal of a square matrix An×n . The remaining diagonals are called secondary diagonals. (5) Diagonal Matrix All elements outside the main diagonals are equal to zero. 

An×n

      =       

a11

0

0

···

0

a22

0

···

0

0

a33

···

.. .

.. .

.. .

..

0

0

0

···

.

0



  0     0   ..   .    ann

with ai j ̸= 0 for i = j The elements ai j ̸= 0 form the main diagonal of the matrix An×n .

5.1 Fundamental Terms

93

(6) Identity Matrix ’I’ All elements of the main diagonal are one, all others are equal to zero.   1 0 ··· 0    0 1 ··· 0    In×n =  . . ..   .. .. .   0 0 ··· 1

Unit vectors are equivalent to vectors that consist of exactly a single one and otherwise zeros. Example:   1   0     i1 =  0  . . . 0

  0   1     i2 =  0  . . .

···

  0   0     in =  ...      0

0

1

(7) Triangular Matrix All elements (of a square matrix An×n ) on one side of the main diagonal are equal to zero. Examples: 

15

√

 0 0 A4×4 =   0 0 00 

100

7 −1



 8 π   2 e 0

= b

upper triangular matrix

3



   A3×3 =  0 2 0 104

= b

lower triangular matrix

5 Linear Algebra

94 (8) Symmetric Matrix

The row elements above the main diagonal are equivalent to the column elements below the main diagonal ⇒ A = A′ . Example:





 3 5 2  ′  A=  5 7 6 =A   2 6 9

5.2 Operations with Matrices 5.2.1 Addition of Matrices Two matrices of the same order are added together by adding the elements in the same position to each other. A2×2 + C2×3

⇒

Addition not possible due to unequal order.

A2×2 + B2×2

⇒

Addition possible here because of same order.



a11 a12 · · · a1n

 a  21 Am×n =  .  ..  am1

 a2n   ..  .   · · · amn

a22 · · · .. . am2 





b11 b12 · · · b1n

 b  21 Bm×n =  .  ..  bm1

 b2n   ..  .   · · · bmn

b22 · · · .. . bm2

a11 + b11 a12 + b12 · · · a1n + b1n





   a +b a +b ··· a +b  21 22 22 2n 2n   21 Am×n + Bm×n =   .. .. ..   . . .   am1 + bm1 am2 + bm2 · · · amn + bmn

95

5.2 Operations with Matrices Examples:

A=

235

! B=

!

0 −7 1

147 1 5 5

A+B =

−1 2 0

! A−B =

1 −3 8

3 1 5

!

1 11 6

Graphical illustration of the addition of two vectors in the two-dimensional space S2 :

a=

5

!

6

c=

a + b′

 b = 11

 2

1×2

2×1

=

5 6

! +

11 2

! =

16 8

!

5 Linear Algebra

96 Laws of Addition of Matrices

The following laws apply to matrices of the same order: (1) A + B = B + A

(commutative law)

(2) A + B +C = (A + B) +C = A + (B +C)

(associative law)

(3) A ± 0 = A (4) if

A+B = 0

⇒

B = −A

with

A = [ai j ]m×n

(5) (A + B)′ = A′ + B′

5.2.2 Multiplication of Matrices 5.2.2.1 Multiplication of a Matrix with a Scalar If there is

k = [k]1×1

with

k∈R

and

A = [ai j ]m×n

with

ai j ∈ R; i = 1, ..., m; j = 1, ..., n



a11 · · ·  . k · Am×n = k ·   ..

then:

 a1n ..   . 

am1 · · · amn Examples:

(1)

    5 10        2· = 4  8  6 12

m×n

 =

 k · a11 · · · k · a1n  . ..   .  .   . k · am1 · · · k · amn

97

5.2 Operations with Matrices  (2)

100





77 0 0



       77 · I3×3 = 77 ·   0 1 0  =  0 77 0  001 0 0 77  9 7 3 −  11 11 11   = 1 ·  1 8 5  11 − 11 11 11 

(3)

−9 7 3

!

1 −8 5

Graphical illustration of the multiplication of a vector a by a scalar k, (k ∈ R):

a=

4

!

2

2·a =

8

! ;

−0.5 · a =

−2 −1

4

k>1

= b

dilation

k 3) Determinants of (n × n)-Triangular Matrices Analogous to (b) Determinants of 3 × 3-Matrices, it is also possible to apply the expansion along a row or along a column according to Laplace to (square) matrices of higher order. 

3 4

7

1



   0 2 1 −1  =  det    4 5 −2 0  1 2

1

3

by means of the expansion e.g. along the 1st row 

2 = +3  5 2

1 −2 1

  −1 0  −4 4 0 3 1

1 −2 1

  −1 0  +7 4 0 3 1

2 5 2

  −1 0  −1 4 0 3 1

= +3 · (−36) − 4 · (−18) + 7 · (−27) − 1 · (−9) = −216 

3 4

7

1



   0 2 1 −1   = det    4 5 −2 0  1 2

1

3

by means of the expansion e.g. along the 2nd column

2 5 2

 1 −2  = 1

123

5.5 The Determinant of a Matrix 

0 = −4  4 1

1 −2 1

  −1 3 0 +2 4 3 1

7 −2 1

  1 3 0 −5 0 3 1

7 1 1

  1 3 −1  + 2  0 3 4

7 1 −2

 1 −1  = 0

= −4(−18) + 2(−96) − 5 · 4 + 2 · (−38) = −216 

4 15 −27 −13



 0 det   0

2

−8

0

5

 46  =  107 

0

0

0

−7

by means of the expansion e.g. along the 3rd row       15 −27 −13 4 15 −13 4 15 −27            = +0   2 −8 46  − 0  0 2 46  − 107  0 2 −8  = 0 0 −7 0 0 −7 0 0 0 = +0 · 462 − 0 · 224 + 5 · (−56) − 107 · 0 = −20 As an alternative, the determinant of an upper or lower triangular matrix can be formed by the product of the elements of the main diagonals. 

a11 a12 a13 · · · a1n

    det    



 a22 a23 · · · a2n   n  a33 · · · a3n  = a11 · a22 · ... · ann = ∏ aii i=1 ..   .  0

ann

Example:   4 15 −27 −13    2 −8 46    = 4 · 2 · 5 · (−7) = −280 det    5 107  0

−7

5 Linear Algebra

124 Determinants of 4 × 4-Matrices Example:   3 4 7 1    0 2 1 −1   A=    4 5 −2 0  1 2

1

3

Formation of an upper triangular matrix   3 4 7 1   0 2 1 −1    ⇒ 9  0 0 − 67 − 6 6 0 0

0

3.2

  67 det A = 3 · 2 · − · 3.223 = −216 6 → simplest method with 4 × 4-matrices

5.5.3 Characteristics of Determinants Let A be an (n × n)-matrix, then: (1) det A = det A′ Example:

det

1 2

! = det

0 3

1 0

! =3

2 3

(2) Swapping two rows/columns changes the sign of the determinants and thus the result.

Example:

det

1 2 0 3

! =3

det

2 1 3 0

! = −3

5.6 The Adjoint of a Matrix

125

(3) The row/column vectors of the matrix A are linearly dependent, if: det A = 0, i.e. A is singular → no inverse formation possible.

Example:

2 4

! = 0 → linearly dependent

1 2 (4) detA = 0, if all elements of a row or a column are zero. (5) For two (n × n)-matrices, A, B, the following applies: det(A · B) = detA · detB However, in general: det(A + B) ̸= detA + detB

5.6 The Adjoint of a Matrix 5.6.1 Definition The adjoint of a matrix is the transpose of the cofactor matrix. Multiplying the minor detAi j by the factor (−1)i+ j results in the cofactor αi j of the element ai j . If the cofactors αi j , with i, j = 1, ..., n, are combined into a matrix, the cofactor matrix [αi j ]n×n is formed. If the cofactor matrix [αi j ]n×n is transposed, the adjoints are finally formed. [αi j ]′n×n = Aad of the original matrix A.  Example:

324



   A= 1 0 2 375

5 Linear Algebra

126 ⇒ A3×3 exist 32 = 9 sub-determinants (= minors).

detA11 = det

0 2

! = 0 · 5 − 2 · 7 = −14 ⇒ α11 = (−1)1+1 · (−14) = −14

7 5

detA12 = det

1 2

! = 1 · 5 − 2 · 3 = −1 ⇒ α12 = (−1)1+2 · (−1) = 1

3 5

detA13 = det

1 0

! = 1 · 7 − 0 · 3 = 7 ⇒ α13 = (−1)1+3 · 7 = 7

3 7

detA21 = det

2 4

! = 2 · 5 − 4 · 7 = −18 ⇒ α21 = (−1)2+1 · (−18) = 18

7 5

detA22 = det

3 4

! = 3 · 5 − 4 · 3 = 3 ⇒ α22 = (−1)2+2 · 3 = 3

3 5

detA23 = det

3 2

! = 3 · 7 − 2 · 3 = 15 ⇒ α23 = (−1)2+3 · 15 = −15

3 7

detA31 = det

2 4

! = 2 · 2 − 4 · 0 = 4 ⇒ α31 = (−1)3+1 · 4 = 4

0 2

detA32 = det

3 4

! = 3 · 2 − 4 · 1 = 2 ⇒ α32 = (−1)3+2 · 2 = −2

1 2

detA33 = det

3 2 1 0

! = 3 · 0 − 2 · 1 = −2 ⇒ α33 = (−1)3+3 · (−2) = −2

127

5.6 The Adjoint of a Matrix 

−14 1

7



 ⇒ |αi j |3×3 =   18

 3 −15   = cofactor matrix 4 −2 −2 

−14 18

4



 ⇒ Aad = |αi j |′3×3 =   1

 3 −2   = adjoint 7 −15 −2

5.6.2 Determination of the Inverse with the Usage of the Adjoint The following is valid: A−1 =

1 1 · Aad = · [αi j ]′n×n detA detA

Remark: The matrix A is only regular (invertible), if det A ̸= 0.

Example: 

3 2 4



   A= 1 0 2 3 7 5 A−1 =? The following is valid: A−1 =

1 · Aad detA

5 Linear Algebra

128 

−14 1

7



 [αi j ]3×3 =   18

 3 −15   = cofactor matrix 4 −2 −2 ⇒ see example in chapter 5.6.1.



−14 18

4



  ⇒ Aad =  3 −2   1  = adjoint 7 −15 −2 detA =? Calculation e.g. with the aid of Sarrus’ Rule (see chapter 5.5.2). 

3 2 4



3 2    A= 1 0 2 1 0 3 7 5 3 7 det A = 3 · 0 · 5 + 2 · 2 · 3 + 4 · 1 · 7 − 4 · 0 · 3 − 3 · 2 · 7 − 2 · 1 · 5 = −12  A−1 =

−14 18

4



 1 1  = · Aad = − ·  1 3 −2   detA 12 7 −15 −2

  14   12   = − 1  12    7 − 12

 −

18 12

−

3 12

15 12



4   7    12    6    2   = − 1  12    12   2   7 − 12 12

−

 −

3 2

−

1 4

5 4

1 −  3   1   6    1  6

Chapter 6

Combinatorics 6.1 Introduction A basic task of combinatorics is to determine the number of possible arrangements (permutations) for a (basic) population of N different elements e1 , e2 , . . . , eN . Example: A (basic) population of N = 3 elements e1 , e2 , e3 results in six different arrangements:   e1 e2 e3      e1 e3 e2      e2 e1 e3 ⇒ 3! = 1 · 2 · 3 = 6  e2 e3 e1      e3 e1 e2      e3 e2 e1

In general: for N different elements there are N! arrangements (= so-called permutations). N! = 1 · 2 · 3 · 4 · ... · (N − 1) · N

(N!, read: "N factorial")

Remark: 0! = 1 Another important task of combinatorics is the determination of possible arrangements when selecting n elements from a basic population of N elements.

© Springer-Verlag GmbH Germany, part of Springer Nature 2023 F. W. Peren, Math for Business and Economics, https://doi.org/10.1007/978-3-662-66975-4_6

129

6 Combinatorics

130 Urn Model: From an urn with a total of N balls, n balls are drawn. Without repetition of elements:

in each arrangement, each element occurs once at most; sampling without replacement

With repetition of elements:

at least one element can occur multiple times; sampling with replacement

Order is relevant:

the swapping of elements within an arrangement results in a new arrangement (so-called variation)

Order is irrelevant:

the swapping of elements within an arrangement does not result in a new arrangement (so-called combination)

Examples: With each N = 3 and n = 2 elements (1) possible arrangements without repetition of single elements (a) order is relevant:   e1 e2      e1 e3      N! 3! 6 e2 e1 ⇒ = = = 6 variations  (N − n)! (3 − 2)! 1 e2 e3      e3 e1      e3 e2

6.1 Introduction (b) order is irrelevant:     e1 e2 = e2 e1   N  N! 3 3! 6 e1 e3 = e3 e1 ⇒ n = n!(N − n)! = 2 = 2!1! = 2 · 1   e2 e3 = e3 e2 = 3 combinations   N read: "N choose n" (binomial coefficient) n Remark:     N N = =1 0 N     N N = =N 1 N −1     N N = n N −n

(2) possible arrangements with repetition of single elements (a) order is relevant:   e1 e1      e2 e1      e3 e1       e1 e2   n 2 e2 e2 ⇒ N = 3 = 9 variations   e3 e2       e1 e3      e2 e3      e3 e3

131

6 Combinatorics

132

(b) order is irrelevant:   e1 e1      e1 e2      N + n − 1 3 + 2 − 1 4 24 4! e1 e3 ⇒ = = = =  n 2 2 2!(4 − 2)! 2 ·2 e2 e2      e2 e3      e3 e3 = 6 combinations

6.2 Permutations

133

6.2 Permutations Definition: A permutation P of N different elements corresponds to the number of possible arrangements in a (full) survey of all elements. Here it can be distinguished whether an element exists only once in the basic population (without repetition) or whether an element occurs several times and is therefore not distinguishable (with repetition).

without repetition

with repetition

Every element occurs exactly once per arrangement.

The i th element occurs multiple times, i.e. repeatedly.

Pw/o.rep. = N!

Pw/rep. =

Example:

Example:

elements: e1 , e2 ⇒ N = 2

elements: e1 , e2 ⇒ ne1 = 2; ne2 = 1

Pw/o.rep. = 2! = 2

Pw/rep. =

namely e1 e2 ; e2 e1

namely e1 e1 e2 ; e1 e2 e1 ; e2 e1 e1

N! n1 ! · n2 ! · ... · nk !

(2 + 1)! 1 · 2 · 3 = =3 2! · 1! 2·1·1

6 Combinatorics

134 Examples: Permutation without Repetition

Four horses compete in a horse race. How many possibilities are there for the horses to reach the finish line in different orders? Solution: Pw/o.rep. = N! = 4! = 4 · 3 · 2 · 1 = 24

How many ways can five women and three men pass through a revolving door? Solution: Pw/o.rep. = N! = 8! = 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 40, 320

There are four German and three English books on a bookshelf. The German books should be placed on the left side of the bookshelf and the English books on the right side of the bookshelf. How many ways are possible to arrange the books? Solution: Pw/o.rep. = N = N1 ! · N2 ! = 4! · 3! = 4 · 3 · 2 · 1 · 3 · 2 · 1 = 144

Permutation with Repetition In an urn there are two red and three white balls. How many possibilities are there to put them in order? Solution: Pw/rep. =

N! 5! 5·4·3·2·1 = = = 10 n1 ! · n2 ! · ... · nk ! 2! · 3! (2 · 1) · (3 · 2 · 1)

How many possibilities are there to arrange the individual letters of the word MISSISSIPPI?

135

6.3 Variations Solution: in total 11 letters ⇒ N! = 11 1xM, 4xI, 4xS, 2xP Pw/rep. =

⇒ n1 = 1;

n2 = 4;

n3 = 4;

n4 = 2

11! 39, 916, 800 N! = = = 34, 650 n1 ! · n2 ! · ... · nk ! 1! · 4! · 4! · 2! 1, 152

6.3 Variations Definition: A variation V of n elements from a basic population of N different elements is equivalent to the number of possible arrangements, if the order of the elements in the arrangement is relevant. Here it can be distinguished whether elements occur only once (without repetition) or whether they occur multiple times (with repetition).

without repetition Vw/o.rep. =

Example:

Vw/o.rep. =

N! (N − n)! elements: e1 , e2 , e3 3! =6 (3 − 2)!

namely e1 e2 , e1 e3 , e2 e1 , e2 e3 , e3 e1 , e3 e2

with repetition Vw/rep. = N n

⇒ N = 3; n = 2

Vw/rep. = 32 = 9 namely e1 e1 , e2 e1 , e3 e1 , e1 e2 , e2 e2 , e3 e2 , e1 e3 , e2 e3 , e3 e3

6 Combinatorics

136 Examples: Variation without Repetition

Ten cars participate in a car race. How many possibilities are there to fill the first three places, respecting the order? Solution: Vw/o.rep. =

N! 10! 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = = = (N − n)! (10 − 3)! 7·6·5·4·3·2·1

= 720

Variation with Repetition A bicycle lock uses a four-digit code consisting of the numbers 0 to 9. How many possibilities are there if the individual digits may occur several times? Solution: Vw/rep. = N n = 104 = 10, 000

6.4 Combinations Definition: A combination C of n elements from a basic population of N different elements is meant to be the number of possible arrangements, if the order of the elements in the arrangement is irrelevant. Here it can be distinguished whether the elements occur only once (without repetition) or whether they can occur multiple times (with repetition).

6.4 Combinations

without repetition   N n N! = n! · (N − n)!

Cw/o.rep. =

Example:

elements: e1 , e2 , e3

137

with repetition 

 N +n−1 n (N + n − 1)! = (N − 1)! · n!

Cw/rep. =

⇒ N = 3; n = 2

  3 3! Cw/o.rep. = = =3 2 2! · (3 − 2)!



   3+2−1 4 Cw/rep. = = 2 2 4! = =6 2! · (4 − 2)!

namely e1 e2 , e1 e3 , e2 e3

namely e1 e1 , e1 e2 , e1 e3 , e2 e2 , e2 e3 , e3 e3

Examples: Combination without Repetition In a lotto 6 out of 49, exactly six of the 49 numbers should be ticked. How many possibilities are there if the order is not taken into regard and a drawn number may only occur once?     N 49 49! 49! Solution: Cw/o.rep. = = = = = 13, 983, 816 n 6 6! · (49 − 6)! 6! · 43!

For a project, a company wants to put together a team of three employees. How many possibilities are there to form a team when twelve employees are available?     N 12 12! = 220 Solution: Cw/o.rep. = = = 3! · (12 − 3)! n 3

6 Combinatorics

138

In a lecture hall there are nine lamps that can be switched on and off independently of each other. How many possibilities are there if a minimum of six lamps must be lit? Solution: minimum of 6 lamps ⇒ exactly 6, 7, 8 or 9 lamps are lit N = 9;

n1 = 6,

n2 = 7,

n3 = 8,

n4 = 9

          N 9 9 9 9 Cw/o.rep. = ⇒ + + + n 6 7 8 9 = 84 + 36 + 9 + 1 = 130

Combination with Repetition Example 1: From an urn with six different coloured balls, four balls are to be drawn and put back (with repetition). How many possibilities are there if the order is disregarded? Solution: Cw/rep. =

(N + n − 1)! (6 + 4 − 1)! 9! = = = 126 (N − 1)! · n! (6 − 1)! · 4! 5! · 4!

Sweets Ltd. produces candies in the flavours apple, orange, banana, pineapple and blueberry. How many possible candy mixtures are there, if 15 candies fit into one bag and the candies are filled into the bags randomly? Solution: Cw/rep. =

19! (N + n − 1)! (15 + 5 − 1)! = = = 11, 628 (N − 1)! · n! (15 − 1)! · 5! 14! · 5!

Example 2: A jewelery manufacturer produces multicolored pearl necklaces using seven different colors. On one chain 40 pearls are threaded.

6.4 Combinations

139

How many pearl combinations are possible if the pearls are threaded onto the chain purely at random and the individual colors are allowed to repeat? Solution: Cw/rep..: N = 7, n = 40, 

N +n−1 n



 =

i.e. here N < n

   7 + 40 − 1 46 = = 46 nCr -button 40 = 9, 366, 819 40 40

There are 9, 366, 819 pearl combinations for this necklace.

Overview of Basic Combinatorial Formulas:

Selection from a basic population → partial survey n < N

repetition order order relevant → variation order irrelevant → combination

without rep. of single elements

with rep. of single elements

N! (N − n)!

Nn

  N n



 N +n−1 n

All elements are considered → complete survey n = N

permutation

without rep. of single elements

with rep. of single elements

N!

N! n1 ! · n2 ! · ... · nk !

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6 Combinatorics

Chapter 7

Financial Mathematics 7.1 Calculation of Interest 7.1.1 Fundamental Terms

Interest z

interest is the charge for a loaned capital. - debit interest is interest that must be paid. - credit interest is interest that is received.

Interest Rate i

determines what percentage of the initial capital is to be paid at the end of an interest period on the initial capital. The interest rates are classified as follows: - according to the length of the interest period: • annual interest rate (year) • interest rate during the year (fraction of a year, e.g. quarter) - according to the calculated reference value: • interest in arrears (initial capital) • advance interest (final capital) Remark: The standard case is an annual interest rate in arrears.

© Springer-Verlag GmbH Germany, part of Springer Nature 2023 F. W. Peren, Math for Business and Economics, https://doi.org/10.1007/978-3-662-66975-4_7

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7.1.2 Annual Interest 7.1.2.1 Simple Interest Calculation

Interest Factor q

q = 1+i with i = interest rate in decimal notation (e.g. 0.01 for 1%)

Interest Period

the period between two interest payments

Initial Capital C0

capital at the beginning of the period (also called cash value or present value)

Final Capital Cn

capital after the n th (interest) period (at the end of the period)

n

is the period measured in years

t

is the period measured in days

The interest must always be calculated from the initial capital C0 , i.e. the annual interest due always remains the same.

Interest

z1, 2, 3, ..., n = C0 · i z = C0 · n · i (for the period n)

Interest Portion z =

C0 · i (per month) 12

z = C0 · Final Capital

t ·i (for t days) 360

Cn = C0 · (1 + n · i)

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7.1 Calculation of Interest

Cn = C0 + n · z1, 2, 3, ..., n = C0 + n ·C0 · i

Commercial Interest Formula:   t ·i Cn = C0 · 1 + = C0 + z 360 The equation Cn = C0 · (1 + n · i) forms the basis for the calculation of C0 , i and n. By solving it accordingly, the following equations are obtained:

Initial Capital Interest Rate Period

Cn 1+n·i   1 Cn i= · −1 n C0   1 Cn −1 n= · i C0

C0 =

Example: Final Capital

C0 = $2, 000; t = 200 days; interest rate in percentage = 10 % 

 200 · 0.1 = $111.11 360   200 Cn = $2, 000 · 1 + · 0.1 = $2, 000 + $111.11 360 z = $2, 000 ·

Cn = $2, 111.11

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Initial Capital

Cn = $15, 000; n = 8 years; interest rate in percentage = 5.3 % C0 =

$15, 000 1 + 8 · 0.053

C0 = $10, 533.71 Interest Rate

C0 = $840; Cn = $1, 070; n = 4 years interest rate =

  1 1, 070 · −1 4 840

interest rate = 0.0685 interest rate in percentage = 6.85 % Period

C0 = $5, 000; Cn = $7, 000; interest rate in percentage = 5 %   $7, 000 1 · − 1 = 8 years n= 0.05 $5, 000

7.1.2.2 Compound Computation of Interest Interest claims that arise during the period are added to the interestbearing capital at the end of the year. In the following interest periods, the interest of the previous interest periods is also included.

Final Capital

C1 = C0 · (1 + i) C2 = C0 · (1 + i)2

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7.1 Calculation of Interest

C3 = C0 · (1 + i)3 .. . Cn = C0 · (1 + i)n = C0 · qn

Initial Capital

C0 = Cn · (1 + i)−n = r n

Cn = Cn · q−n qn

Cn −1 C0

Interest Rate

i=

Period

Cn = C0 · (1 + i)n ⇔ log(1 + i)n = log(Cn ) − log(C0 ) ⇔ n · log(1 + i) = log(Cn ) − log(C0 ) ⇔n =

log(Cn ) − log(C0 ) log(Cn ) − log(C0 ) = log(1 + i) log(q)

Example: Final Capital

C0 = $12, 500; n = 6 years; interest rate in percentage = 4 % C6 = $12, 500 · (1 + 0.04)6 = $12, 500 · 1.046 C6 = $15, 816.49

Initial Capital

Cn = $2, 500; n = 7 years;

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interest rate in percentage = 5 % C0 =

$2, 500 = $2, 500 · 1.057 (1 + 0.05)7

C0 = $1, 776.70 Interest Rate

C0 = $2, 000; Cn = $4, 000; n = 8 years s $4, 000 − 1 = 0.091 i= 8 $2, 000

Period

C0 = $9, 050; Cn = $11, 000; interest rate in percentage = 3 % log($11, 000) − log($9, 050) 0.08474 = log(1 + 0.03) 0.01284 n = 6.6 years

7.1.2.3 Composite Interest The composite interest calculation is important for fractional periods (e.g. 1 year + 25 days). It represents an addition of the simple interest calculation and the exponential interest calculation. If the interest period covers one year, exponential interest is calculated for full years, since exponential interest is more profitable than linear interest from a full interest period onwards. If the time frame includes less than one interest period (during the year), interest is calculated simply since the linear interest calculation generates more revenue than the exponential interest calculation if interest is only calculated for part of a full period.

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The graph of the linear versus exponential interest rate shows that the straight line is above the exponential function until the first full interest period is reached, since the exponential function has a flatter slope at the beginning. They intersect at t =1 period (here t =1 year). From this point on, the exponential interest rate generates a higher return than the linear interest rate. n = n1 + n2

with

n

total period in years with annual interest periods

n1

period of a first partial period within the entire interest period n, measured in complete (full) years

n2

period of a second partial period (remaining period) within the entire interest period n, which corresponds to the remaining (less than one year) period (n − n1 )

Example: Final Capital C0 = $5, 000; interest rate in percentage = 3.9 % p.a.; from 12.02.2003 to 20.08.2010 February in the years 2004 and 2008 includes 29 days.

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a) 30E/360 ISDA (German interest rate method) The German interest method stipulates that each month is calculated with 30 interest days and a full year with 360 interest days. This means that months that lie entirely between the starting and the ending date of the interest payment period are each counted as 30 days, regardless of the actual number of days they have. If a month has 31 days, the 31st calender day is not an interest day. If the start or the end of the period falls on the 31st of a month, it is treated as the 30th calender day. If the transaction ends on the 28th of February or, in a leap year, on the 29th of February, interest is only calcualted up to this date. If, on the other hand, the transaction goes beyond February, February is treated like every month with 30 days. In this example the 20th of August is not calculated, since according to the commercial interest method the last day of savings deposits on the German capital market does not bear interest (§§ 187, 188 BGB). The first day of investment earns interest, the last day of investment does not. 10.01.2001 to 10.03.2001 → 21 + 30 + 9 = 60 days 28.02.2001 to 10.03.2001 → 3 + 9 = 12 days 10.01.2001 to 28.02.2001 (not a leap year) → 21 + 27 = 48 days 10.01.2000 to 29.02.2000 (leap year) → 21 + 28 = 49 days 10.01.2000 to 01.03.2000 (leap year) → 21 + 30 = 51 days Months are always counted as 30 days, regardless of the actual number of days. The 31st day of a month is not taken into account. If the end or the beginning of the interest period falls on the 31st day of a month, this is not taken into account. The year always has 12 x 30 = 360 days. 10.01.2001 to 31.03.2001 → 21 + 30 + 29 = 80 days 31.01.2001 to 31.03.2001 → 1 + 30 + 29 = 60 days 30.01.2001 to 31.03.2001 → 1 + 30 + 29 = 60 days

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149

    ∆t1 ∆t2 n Cn = C0 · 1 + · i · (1 + i) · 1 + ·i 360 360

Cn

Final Capital

C0

Initial Capital

∆t1

Period from capital contribution to 1st interest payment

∆t2

Period from last interest payment to end of capital investment

i

Interest Rate p.a.

n

Period measured in years

  319 Cn = $5, 000 · 1 + · 0.039 · (1 + 0.039)6 · 360   229 · 1+ · 0.039 360 Cn = $6, 669.00

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Day of deposit → End of year Remaining period 1 {z } | simple interest

19 30 · 10 + 360 360 319 = = n21 360

Period n-years

Interest period {z } |

compound computation of interest

6 years = n1

End of year → Day of payout Remaining period 2 {z } | simple interest

19 30 · 7 + 360 360 229 = = n22 360

10 months + 19 days

7 months + 19 days

Feb. 12th included

Aug. 20th not included

b) 30E/360 ICMA (U.S. interest rate method) The method is similar to the German commercial interest method, as the interest months are set at 30 days and the interest year at 360 days. The exception is February, which is set to the exact calendar date of 28 or 29 days, provided that the start or end of the period falls on these days. The base year, like the interest month and interest year, is set at 360 days regardless of the number of actual days. The first day of investment does not earn interest, the last day of investment does. 10.01.2001 to 10.03.2001 → 20 + 30 + 10 = 60 days 28.02.2001 (not a leap year) to 10.03.2001 → 0 + 10 = 10 days 10.01.2001 to 28.02.2001 (not a leap year) → 20 + 28 = 48 days 10.01.2000 to 29.02.2000 (leap year) → 20 + 29 = 49 days 10.01.2000 to 01.03.2000 → 20 + 30 + 1 = 51 days The 31st calendar day of the month counts, provided that the investment ends on this day and the interest period does not begin on the 30th or 31st of another month.

7.1 Calculation of Interest 10.01.2001 to 31.03.2001 → 20 + 30 + 31 = 81 days 31.01.2001 to 31.03.2001 → 0 + 30 + 30 = 60 days 30.01.2001 to 31.03.2001 → 0 + 30 + 30 = 60 days

    ∆t1 ∆t2 Cn = C0 · 1 + · i · (1 + i)n · 1 + ·i 360 360 For explanation of parameters see a).

  318 Cn = $5, 000 · 1 + · 0.039 · (1 + 0.039)6 · 360   230 · 1+ · 0.039 360 Cn = $6, 669.01

151

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Day of deposit → End of year Remaining period 1 {z } | simple interest

18 30 · 10 + 360 360 318 = = n21 360

Period n-years

Interest period {z } |

compound computation of interest

6 years = n1

End of year → Day of payout Remaining period 2 {z } | simple interest

20 30 · 7 + 360 360 230 = = n22 360

10 months + 18 days

7 months + 20 days

Feb. 12th not included

Aug. 20th included

With all ACT methods, the interest days are determined exactly to the calendar. Consequently, individual months are calculated with 30 or 31 interest days, or February with 28 or 29 interest days, depending on their actual number of days. Depending on the type of investment, interest is calculated either on the first or the last day of investment.

c) Actual/360 (Euro interest rate method) Under the Actual/360 method, the interest days are divided by 360 to determine the proportion of the nominal annual interest rate. This results in 365 interest days for a full year or 366 interest days in a leap year. In the Euro interest rate method, interest is paid on the first day of investment; no interest is paid on the last day of investment.

7.1 Calculation of Interest

153

  t2 365 ∆t1 · i · 1+ ·i · Cn = C0 · 1 + 360 360 

 t3   366 ∆t4 · 1+ ·i · 1+ ·i 360 360

Cn

Final Capital

C0

Initial Capital

∆t1

Period from capital contribution to 1st interest payment

∆t4

Period from last interest payment to end of capital investment

t2

Number of full years that are no leap years

t3

Number of full leap years

i

Interest Rate p.a.

n

Period measured in years

   4 323 365 Cn = $5, 000 · 1 + · 0.039 · 1 + · 0.039 · 360 360  2   366 231 · 1+ · 0.039 · 1 + · 0.039 360 360 Cn = $6, 695.50

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Day of deposit → End of year

Period n-years

Remaining period 1 {z } | simple interest

4 · 30 6 · 31 17 + + 360 360 360 323 = = n21 360

Interest period {z } |

compound computation of interest

4 years with 365 days per year = n11 2 years with 366 days per year = n12

End of year → Day of payout Remaining period 2 {z } | simple interest

2 · 30 4 · 31 28 + + + 360 360 360 231 19 + = = n22 360 360

10 months + 17 days

7 months + 19 days

Feb. 12th included

Aug. 8th not included

d) Actual/360 (French interest rate method) The only difference between the French interest rate method compared to the Euro interest rate method is that the first day of investment does not earn interest, but the last day of investment does.

   t2 ∆t1 365 Cn = C0 · 1 + · i · 1+ ·i · 360 360  t3   366 ∆t4 · 1+ ·i · 1+ ·i 360 360 For explanation of parameters see c).

155

7.1 Calculation of Interest   4  322 365 · 0.039 · 1 + · 0.039 · Cn = $5, 000 · 1 + 360 360  2   366 232 · 1+ · 0.039 · 1 + · 0.039 360 360 Cn = $6, 695.51

Day of deposit → End of year Remaining period 1 | {z } simple interest

16 4 · 30 6 · 31 + + 360 360 360 322 = = n21 360

Period n-years

Interest period {z } |

compound computation of interest

4 years with 365 days per year = n11 2 years with 366 days per year = n12

End of year → Day of payout Remaining period 2 | {z } simple interest

28 2 · 30 4 · 31 + + + 360 360 360 20 232 + = = n22 360 360

10 months + 16 days

7 months + 20 days

Feb. 12th not included

Aug. 20th included

e) Actual/365 Fixed (English interest rate method) This method involves dividing the interest days by 365 to determine the share of the nominal annual interest rate. This is the only difference from the Actual/360 method. No interest is calculated for the first day of investment while it is calculated for the last day of investment.

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  t2 ∆t1 365 Cn = C0 · 1 + · i · 1+ ·i · 360 360 

 t3   366 ∆t4 · 1+ ·i · 1+ ·i 360 360 For explanation of parameters see c).    4 322 365 Cn = $5, 000 · 1 + · 0.039 · 1 + · 0.039 · 365 365 2    232 366 · 1+ · 0.039 · 1 + · 0.039 365 365 Cn = $6, 669.26

Day of deposit → End of year Remaining period 1 {z } | simple interest

16 4 · 30 6 · 31 + + 365 365 365 322 = = n21 365

Period n-years

Interest period {z } |

compound computation of interest

4 years with 365 days per year = n11 2 years with 366 days per year = n12

End of year → Day of payout Remaining period 2 {z } | simple interest

28 2 · 30 4 · 31 + + + 365 365 365 20 232 + = = n22 365 365

10 months + 16 days

7 months + 20 days

Feb. 12th not included

Aug. 20th included

7.1 Calculation of Interest

157

f) Actual/Actual ICMA The day-specific interest method provides that both the number of interest days and the length of the base year are always determined to the calendar. This results in 365 interest days for a full year or 366 interest days for a leap year.

  t2 ∆t1 365 Cn = C0 · 1 + · i · 1+ ·i · 360 360 

 t3   366 ∆t4 · 1+ ·i · 1+ ·i 360 360 For explanation of parameters see c).

   4 365 322 Cn = $5, 000 · 1 + · 0.039 · 1 + · 0.039 · 365 365  2   366 232 · 1+ · 0.039 · 1 + · 0.039 366 365 Cn = $6, 667.89

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Day of deposit → End of year Remaining period 1 {z } | simple interest

4 · 30 6 · 31 16 + + 365 365 365 322 = = n21 365

Period n-years

Interest period {z } |

compound computation of interest

4 years with 365 days per year = n11 2 years with 366 days per year = n12

End of year → Day of payout Remaining period 2 {z } | simple interest

2 · 30 4 · 31 28 + + + 365 365 365 232 20 + = = n22 365 365

10 months + 16 days

7 months + 20 days

Feb. 12th not included

Aug. 20th included

7.1.3 Interest During the Period (Sub-annual) parts of a year, usually a calendar year, are defined as interest period(s) (semi-annual, quarterly, monthly or daily interest rates).

m

is the number of sub-annual interest periods per year

j

is the relative periodic interest rate linearly distributed over the respective equally long interest periods during the year m j=

i m

Interest during the year is calculated in the same way as the annual interest rate.

159

7.1 Calculation of Interest 7.1.3.1 Simple Interest Calculation (linear) Final Capital

Cn = C0 · (1 + n · i) = C0 · (1 + N · j)

Initial Capital

C0 =

Cn Cn = 1+n·i 1+N · j   1 Cn j= · −1 N C0   1 Cn −1 N= · j C0

Interest Rate Period

Example: C0 = $3, 000; interest rate in percentage= 7 (i = 0.07); N1 = 5 quarters; N2 = 0.3 quarters j=

7 = 1.75 % 4

C5.3 = $3, 000 · (1 + 5.3 · 0.0175) = $3, 278.25 with j = 1.75 %

7.1.3.2 Simple Interest Using the Nominal Annual Interest Rate Nominal Interest Rate

C0 · (1 + n · i) = C0 · (1 + N · j) C0 · (1 + n · i) = C0 · (1 + m · n · j) i = m· j

Final Capital

Cn = C0 · (1 + n · i)

with

N = m·n

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Relative Interest Rate

Cn 1+n·i   1 Cn j= · −1 m · n C0

Period

n=

Example:

C0 = $2, 000; j = 1.25 %; n = 3.5;

Initial Capital

C0 =

  1 Cn · −1 i C0

m = 4, i.e. quarterly interest Nominal Interest Rate

i = 4 · 0.0125 = 0.05

Final Capital

C3.5 = $2, 000 · (1 + 3.5 · 0.05) = $2, 350

7.1.3.3 Compound Interest (exponential)

Final Capital

Cn = C0 · (1 + j)n

Initial Capital

C0 = Cn · (1 + j)−n r

Interest Rate

Period

Cn −1 C0   Cn ln C0 n= ln (1 + j) j=

n

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7.1 Calculation of Interest

Example:

Cn = $20, 000; interest rate in percentage= 7 (i = 0.07); m = 2, i.e. semi-annual interest 7 = 3.5 % 2

Relative Periodic Interest Rate

j=

Initial Capital

C0 = $20, 000 · (1 + 0.035)−15.5 = $11, 734.23 with a relative periodic interest rate of j = 3.5 %

7.1.3.4 Interest with Compound Interest Using a Conforming Annual Interest Rate A so-called conforming (periodic) interest rate icon f orm (hereinafter referred to as icon ) leads by definition to the same result as the annual interest rate i for m interest periods of less than one year.

Conforming Periodic Interest Rate

C0 · (1 + icon )n = C0 · (1 + j)N C0 · (1 + icon )n = C0 · (1 + j)m·n icon = (1 + j)m − 1

Final Capital

Cn = C0 · (1 + icon )n

Initial Capital

C0 = Cn · (1 + icon )−n r

Interest Rate

j=

m·n

Cn −1 C0

with

N = m·n

7 Financial Mathematics

162  Cn ln C0 n= ln(1 + icon ) 

Period Example:

C0 = $4, 000; j = 0.5 %; m = 4; n = 6.5

Conforming Periodic Interest Rate

icon = (1 + 0.005)4 − 1 = 0.02015050063 ≈ 2 %

Final Capital

C0 · (1 + icon )n = C0 · (1 + j)m·n $4, 000 · (1 + 0.02015050063)6.5 = $4, 553.84 $4, 000 · (1 + 0.005)4·6.5 = $4, 553.84

The present example demonstrates that a so-called conforming (periodic) interest rate icon for m interest rates during the year leads by definition to the same result as the (sub-)annual interest rate j.

7.1.3.5 Mixed Interest

Final Capital

Cn = C0 · (1 + j)n1 · (1 + n2 · j)

with

n1 = int(n) n2 = n − n1

int(...) represents the integer function commonly used by pocket calculators. This means that n1 is the largest number to which n1 ≤ n applies. Consequently, n2 is limited to the interval from 0 to 1, n2 ∈ [0, 1].1 1

Cf. Kruschwitz, L. (2018): Finanzmathematik, 6th edition, p. 6.

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163

Cn (1 + j)n1 · (1 + n2 · j)

Initial Capital

C0 =

Interest Rate

Calculation of the zeros of the function f (i) = −Cn +C0 · (1 + j)n1 · (1 + n2 · j)

Period

Example:

  1 Cn n = n1 + · −1 j C0 · (1 + j)n1   Cn ln C0 with n1 = int ln(1 + j) C0 = $10, 000 interest rate in percentage = 5 % p.a. n1 = 12 half-years; n2 = 3 months = 0.5 half-years j=

0.05 = 0.025 2

Cn = $10, 000 · (1 + 0.025)12 · (1 + 0.5 · 0.025) Cn = $13, 616.99

7.1.3.6 Steady Interest Rate The steady interest rate is a special form of interest during the year, in which the number of interest periods m is infinite or converges towards infinity. The duration of an interest period is approaching zero. Interest income is generated in infinitesimally short periods and accumulated to the (respective previous) capital. Capital and interest income are then (immediately) paid interest again (compound interest). Conse-

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quently, for a given nominal interest rate, the return of interest is higher with a steady interest rate than with a discrete interest rate (annual, semi-annual, etc.).

e

Euler’s number (2.71828...)

n

Number (period) of interest in years

i

Interest rate p.a.

Final Capital

    i m·n Cn = lim C0 · 1 + m→∞ m = C0 · ei·n

Initial Capital

C0 = Cn · e−i·n

Interest Rate

i=

ln(Cn ) − ln(C0 ) n

Period

n=

ln(Cn ) − ln(C0 ) i

Example:

C0 = $1, 000 interest rate in percentage = 3.3 % p.a. n = 5.75 years Note: In this example, a so-called conforming (periodic) interest rate icon was used. This by definition leads to the same result as the (sub-)annual interest rate j. Both interest rates were used as an example for semi-annual interest rates.

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7.1 Calculation of Interest

• with semi-annual interest rates using the (sub-)annual interest rate: j=

0.033 = 0.0165 2

C5.75 = $1, 000 · (1 + 0.0165)5.75·2 ≈ $1, 207.08 using the conforming interest rate:  icon =

1+

0.033 2

2 − 1 ≈ 0.033272

C5.75 $ = 1, 000 · (1 + 0.033272)5.75 ≈ $1, 207.08

This example once again demonstrates that a so-called conforming (periodic) interest rate icon with m sub-annual interest rates by definition leads to the same result as the (sub-)annual interest rate j.

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• with quarterly interest rates  icon =

0.033 1+ 4

4 − 1 ≈ 0.033411

C5.75 = $1, 000 · (1 + 0.033411)5.75 ≈ $1, 208.01

• with monthly interest rates  icon =

0.033 1+ 12

12 − 1 ≈ 0.033504

C5.75 = $1, 000 · (1 + 0.033504)5.75 ≈ $1, 208.63

• with daily interest rates a) Actual/360  icon =

0.033 1+ 360

360 − 1 ≈ 0.033549

C5.75 = $1, 000 · (1 + 0.033549)5.75 ≈ $1, 208.94

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167

b) Actual/365, Actual/Actual

icon =

  0.033 365 1+ − 1 ≈ 0.033549 365

C5.75 = $1, 000 · (1 + 0.033549)5.75 ≈ $1, 208.94

c) Actual/Actual (in case of a leap year)

icon =

  0.033 366 1+ − 1 = 0.033549 366

C5.75 = $1, 000 · (1 + 0.033549)5.75 ≈ $1, 208.94

• with steady interest rates C5.75 = $1, 000 · e0.033·5.75 ≈ $1, 208.95

The present example demonstrates that the daily interest rate comes very close to the steady interest rate in the result (capital end value after 5.75 years C5.75 ), which is justified by the relatively short period of 5.75 years. On the other hand, the other periodic differences shown here are also significant, even for this short period.

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7.2 Annual Percentage Rate The annual percentage rate (APR) allows several credit offers with the same fixed interest periods to be compared. When calculating the effective annual interest rate, fees such as processing fees and discounts are included in addition to the nominal annual interest rate. Sometimes the APR corresponds to the nominal APR, the simple-interest rate for a year, and sometimes to the effective APR, the fee and compound interest rate calculated across a year.2

Effective Annual Percentage Rate There is no exact legal definition of the effective APR. It depends on the type of fees inlcuded. The calculation of the APR can also be differentiated into at least three ways depending on if fees are added to the entire amount or treated as a short-term loan due in the first payment: • Calculating the interest rate for each year without considering fees Example: loan: $200; interest rate: 6 % p.a.; unique fee: $20 1.0612 = 2.0122 ≈ 100 % increase

• The origination fees are added on to the balance due; the total amount is treated as the basis for calculating compound interest Example: loan: $200; interest rate: 6 % p.a.; unique fee: $20 $20 = 0.1 $200

0.1 + 0.06 = 0.16

1.1612 = 5.9360 ≈ 500 % increase

• The origination fees are amortisation as a short-term loan. This loan becomes due with the first payment(s), and the unpaid balance

2

Cf. Wikimedia Foundation Inc. (Ed.) (2020): https://en.wikipedia.org/wiki/Annual_percentage_rate#cite_note-9, accessed 9 December 2022.

7.2 Annual Percentage Rate

169

is amortised as a second long-term loan.The additional first payment(s) is intended mainly to pay the commitment fees and interest charges for that portion.3 United States In the U.S., the calculation of APR is directed by the Truth in Lending Act, which is implemented by the Consumer Financial Protection Bureau (CFPB).4 APR is expressed as a periodic interest rate times number of compounding periods during a year (e.g. semi-annual, quarterly, monthly, daily), which is also called the nominal interest rate.5 The APR must include certain non-interest charges and fees. It has to be disclosed to the borrower within three days of applying for a mortgage. In the U.S. a distinction is made between a “close-ended credit” and an “open-ended credit”. Close-ended Credit In the U.S., a close-ended credit is a type of credit where the funds are distributed in full when the loan is terminated and the loan ends. It must be paid back by a specific date, including interest and finance charges. The loan may require the full payment of principal at maturity, or it may require regular principal and interest payments in defined periods. Close-ended credits are mainly used for home mortgages or auto loans.6 For a fixed-rate mortgage, the APR is equal to the internal rate of return, if prepayment and default would be zero.

3

Cf. Wikimedia Foundation Inc. (Ed.) (2020): https://en.wikipedia.org/wiki/Annual_percentage_rate#cite_note-9, accessed 9 December 2022. 4 Cf. Wikimedia Foundation Inc. (Ed.) (2020): https://en.wikipedia.org/wiki/Annual_percentage_rate#cite_note-9, accessed 9 December 2022. 5 Cf. Tucker, W.R. (2000): Effective Interest Rate (EIR). In: Bankakademie Micro Banking Competence Center (Ed.): https://web.archive.org/web/20051103034219/http: //www.uncdf.org/mfdl/readings/EIR_Tucker.pdf, accessed 13 October 2020. 6 Cf. Federal Deposit Insurance Corporation (Ed.) (2014): https://www.fdic.gov/regulations/laws/rules/6500-3550.html, accessed 9 December 2022.

170

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For an adjustable-rate mortgage the APR will also depend on the prospective trajectory of the index rate.7 Open-ended Credit In the U.S., open-ended credit is a preapproved loan between a financial institution and a borrower that may be used up to a certain limit and can subsequently be paid back.8 The preapproved amount will be defined in a formal agreement between the lender and the borrower.9 Open-ended credits are mainly used for credit cards, home equity loans or other lines of credit.10

European Union In the EU a single method of calculating the APR was introduced in 1998 (directive 98/7/EC), whose publication is needed for the major part of loans. Given the enhanced notation of directive 2008/48/EC, the basic equation for the calculation of APR in the EU is:

M

APR

N

APR

∑ Ci (1 + 100 )−ti = ∑ D j (1 + 100 )−s j

i=1

j=1

M

total number of drawdowns paid by the lender

N

total number of repayments paid by the borrower

i

sequence number of a drawdown paid by the lender

j

sequence number of a repayment paid by the borrower

7

Cf. Tucker, W.R. (2000): Effective Interest Rate (EIR). In: Bankakademie Micro Banking Competence Center (Ed.): https://web.archive.org/web/20051103034219/http: //www.uncdf.org/mfdl/readings/EIR_Tucker.pdf, accessed 13 October 2020. 8 Cf. Federal Deposit Insurance Corporation (Ed.) (2009): https://www.fdic.gov/regulations/laws/rules/6500-1650.html#6500226.14, accessed 9 December 2022. 9 Cf. Twin, A. (2019): Open-End Credit. In: Investopedia (Ed.): https://www.investopedia.com/terms/o/openendcredit.asp, accessed 9 December 2022. 10 Cf. Wikimedia Foundation Inc. (Ed.) (2020): https://en.wikipedia.org/wiki/Annual_percentage_rate#cite_note-9, accessed 9 December 2022.

7.2 Annual Percentage Rate

171

Ci

cash flow amount for drawdown number i

Dj

cash flow amount for repayment number j

ti

interval, expressed in years and fractions of a year, between the date of the first drawdown and the date of drawdown i

sj

interval, expressed in years and fractions of a year, between the date of the first drawdown and the date of repayment j - The EU formula makes use of the natural convention that all time intervals in ti and s j are measured relative to the date of the first drawdown, hence t1 = 0. However, any other date could be used without affecting the calculated APR, as long as it is used consistently. - The left side of this equation represents the present value of the drawdowns made by the lender and the right side shows the present value of the repayments made by the borrower. - Neither the amounts nor the periods between transactions are necessarily equal. For the purpose of this computation it is assumed that a year has 365 days (366 days in a leap year), 52 weeks or 12 equal months. - The result must be given with at least one decimal place.

Examples: Example 1: different repayment amounts Amount borrowed: $1, 000 Repayment:

after 3 months: $274 after 6 months: $274 after 12 months: $548

$1, 000 =

1+

$274 (1 + APR 100 )

APR =q 100

3 12

+

$274 (1 + APR 100 )

6 12

+

$548 12

12 (1 + APR 100 )

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172 $1, 000 =

$274 q

1 4

+

$274 q

1 2

+

$548 q1

using numerical solution methods results in: q ≈ 1.1442283 ie f f = q − 1 ≈ 1.1442283 − 1 ≈ 0.1442283 ≈ 14.42 %

Example 2: Discount and interest payments during the year Amount borrowed: $5, 000 (finally due) Discount: 10 % n = 15 months APR = 7.5 % (interest payments at the end and in the middle of each calendar year) Interest payments: after 3 months (31.12.): $93.75 after 9 months (01.07.): $187.50 after 15 months (31.12.): $187.50 $4, 500 =

$93.75 (1 +

1+

3 APR 12 100 )

+

$187.50 (1 +

+

9 APR 12 100 )

$187.50 + $5, 000 15

12 (1 + APR 100 )

APR =q 100

$4, 500 =

$93.75 q

1 4

+

$187.50 q

3 4

+

$5, 187.50 5

q4

using numerical solution methods results in: q ≈ 1.1742722 ie f f = q − 1 ≈ 1.1742722 − 1 ≈ 0.1742722 ≈ 17.43 %

173

7.3 Depreciation

7.3 Depreciation Impairments are recognized in the accounts through depreciation over their economic life. This concerns fixed and current assets. A distinction is made between time depreciation and performance depreciation.

7.3.1 Time Depreciation The acquisition costs or production costs are distributed among the years of the economic life.

7.3.1.1 Linear Depreciation A

original cost or par

n

economic life in years

Qk

amount of depreciation, by which the book value is reduced in the kth year

Rk

book value after k years (with k = 1, 2, 3, ..., n) Rk = A − ∑ Qk

Rn

residual value (salvage value, old value, scrap value) at the end of the economic life

i

depreciation rate

With linear depreciation, the difference between acquisition or production costs and the residual value at the end of the economic life is distributed evenly over the periods of use. It is assumed that the value is consumed evenly over the useful life. The following applies:

Q1 = Q2 = ... = Qn =

A − Rn n

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174 Example:

A company acquires a vehicle with a value of $90,000, assuming an economic life of 9 years. It is also assumed that the vehicle can be sold for $9,000 at the end of its useful life. The company opts for linear depreciation.

A = $90, 000; Rn = $9, 000; n = 9 amount of depreciation:

Q1 =

$90, 000 − $9, 000 = $9, 000 9

depreciation rate:

i=

$9, 000 · 100% = 11.11 % $90, 000 − $9, 000

7.3.1.2 Arithmetic-Degressive Depreciation With arithmetic-degressive depreciation, the annual depreciation amounts are reduced by a constant amount d. Thus, the first years are more heavily burdened than the later ones. This results in an assumption of decreasing depreciation over the economic life.

d

degressive amount

N

total of the years’ numbers of the economic life

Tk

remaining useful life at the beginning of the year after k years

Degressive Amount

d=

original cost − residual value total of the years’ numbers

175

7.3 Depreciation

or d = or d =

A − Rn n · (n + 1) 2

or d =

2 · (A − Rn ) n · (n + 1)

or d =

A − Rn N

Qk = d · Tk

Amount of Depreciation

Example:

A − Rn 1 + 2 + 3 + ... + n

A company acquires a vehicle with a value of $90,000, assuming an economic life of 9 years. It is also assumed that the vehicle can be sold for $9,000 at the end of its useful life. It is assumed that the depreciation will decrease by a constant amount.

A = $90, 000; Rn = $9, 000; n = 9 d=

$90, 000 − $9, 000 = $1, 800 1 + 2 + 3 + 4 + ... + 9

Q1 = $1, 800 · 9 = $16, 200 Q2 = $1, 800 · 8 = $14, 400 Q3 = $1, 800 · 7 = $12, 600 .. . Q9 = $1, 800 · 1 = $1, 800

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176

The example illustrates that the degressive amount of $1,800 corresponds to the amount of depreciation in the last year of the economic life.

7.3.1.3 Geometric-Degressive Depreciation With geometric-degressive depreciation, the annual amounts of depreciation are reduced by the depreciation rate. Determination of the book values Rk and the residual value Rn Beginning of the 1st year A End of the 1st year 2nd

R1 = A − A · i = A · (1 − i)

year

R2 = R1 − R1 · i = R1 · (1 − i) = A · (1 − i)2

End of the 3rd year .. .

R3 = R2 − R2 · i = R2 · (1 − i) = A · (1 − i)3

End of the nth year

Rn = Rn−1 − Rn−1 · i = Rn−1 · (1 − i)

End of the

Rn = A · (1 − i)n Determination of the amounts of depreciation Qk End of the 1st year

Q1 = A · i

End of the

2nd

year

Q2 = R1 · i = A · (1 − i) · i

End of the .. .

3rd

year

Q3 = R2 · i = A · (1 − i)2 · i

End of the nth year

Qn = Rn−1 · i = A · (1 − i)n−1 · i

177

7.3 Depreciation Determination of the depreciation rate i =

p 100

The depreciation rate i is determined by the ratio of the target residual value Rn to the original value A.

⇔ ⇔ ⇔

 p n Rn = A · 1 − 100  Rn p n = 1− A 100 r p n Rn = 1− A 100 r ! Rn p = 100 · 1 − n A

Example:

|÷A √ | n ... | − 1; ·(−100)

A company acquires a vehicle with a value of $90,000, assuming an economic life of 9 years. It is also assumed that the vehicle can be sold for $9,000 at the end of its useful life. The amount of depreciation shall decrease annually by a constant depreciation rate. A = $90, 000; Rn = $9, 000; n = 9 r   $9, 000 ≈ 22.57 % p = 100 · 1 − 9 $90, 000 Q1 = $90, 000.00 · 0.2257 = $20, 313.00 R1 = $90, 000.00 − 20, 313.00 = $69, 687.00 Q2 = $69, 687.00 · 0.2257 = $15, 728.36 R2 = $69, 687.00 − $15, 728.36 = $53, 958.64

7 Financial Mathematics

178 k

Qk in $

Rk in $

1

20,313.00

69,687.00

2

15,728.36

53,958.64

3

12,178.47

41,780.17

4

9,427.78

32,352.39

5

7,301.93

25,050.46

6

5,653.89

19,396.57

7

4,377.81

15,018.76

8

3,389.73

11,629.03

9

2,624.67

9,004.36

Note: Differences are due to rounding errors.

7.3.2 Units of Production Depreciation In accordance with the changing use of the assets, depreciation is made according to the intensity of use. The amount of depreciation for a period depends on the performance consumed in that period. Therefore, there can usually be no uniform trend for the course of the annual amounts of depreciation or a constant rate of depreciation.

PA

total performance of the asset

PPt

performance consumed during the period

Example:

A company acquires a vehicle worth $90,000, assuming a total performance of 300,000 miles. In the first year, the vehicle covers 50,000 miles. It should be depreciated according to consumption. A = $90, 000; PA = 300, 000 miles; PPt = 50, 000 miles

179

7.3 Depreciation Qk =

50, 000 · $90, 000 = $15, 000 300, 000

7.3.3 Extraordinary Depreciation In addition to the previously explained scheduled depreciation methods which record a permanent decline in value, extraordinary depreciation can also be applied. Unscheduled or extraordinary depreciation records impairment losses that are not caused by the planned, assumed use. This is the case, for example, in the event of extraordinary technological progress or in the event of unforeseeable, i.e. unplanned damage to property.

Example:

A company acquires a vehicle valued at $90,000. After the first two years of depreciation, the vehicle has a book value of $72,000. In the third year, the vehicle is involved in an accident and is severely damaged. Despite repairs, the vehicle has lost value as a car that was involved in an accident. An appraiser certifies a current value of $40,000. The regular depreciation is based on an economic life of 9 years and a selling price of $9,000 at the end of the useful life.

Regular amount of depreciation: Q1 =

$90, 000 − $9, 000 = $9, 000 9

Extraordinary amount of depreciation: $72, 000 − $40, 000 = $32, 000

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180

Instead of the scheduled depreciation amount of $9,000, a depreciation amount of $32,000 is applied in the third year of the useful life. This means that the vehicle is in the books of the company with a current (book) value of $40,000.

7.4 Annuity Calculation 7.4.1 Fundamental Terms An annuity r is a recurring payment made or received at regular intervals. The payments can be either deposits or disbursements.

Present Value of Annuity R0

Total value of an annuity at the beginning of the payment period

Amount of Annuity Rn

Total value of an annuity after n years

Annuity r

Regularly paid instalment

Interest Factor q

Annual interest factor q = 1 + i

Accumulation Factor

Accumulates interest on a monetary amount exponentially (with interest and compound interest) over n periods. qn = (1 + i)n

Discount Factor

Discounts a monetary amount exponentially (with interest and compound interest) over n periods.

181

7.4 Annuity Calculation q−n = (1 + i)−n

Annuity Present Value Factor

qn − 1 qn · (q − 1)

=

(1 + i)n − 1 (1 + i)n · i

With the aid of the annuity present value factor, the present value of annuities of uniform (annuity) payments can be determined. (a) (annually) in arrears: R0 = r ·

qn − 1 qn · (q − 1)

= r·

qn − 1 qn · i

(b) (annually) in advance: R0 = r ·

Final Annuity Value Factor

q · (qn − 1) q · (qn − 1) = r· n q · (q − 1) qn · i

(1 + i)n − 1 qn − 1 = q−1 i With the aid of the final annuity value factor, the amount of annuity of uniform (annuity) payments can be determined. (a) (annually) in arrears: Rn = r ·

qn − 1 qn − 1 = r· q−1 i

(b) (annually) in advance: Rn = r ·

q · (qn − 1) q · (qn − 1) = r· q−1 i

7 Financial Mathematics

182 Annuity Factor

The annuity factor distributes a fixed amount of money at equal annuities A, taking into account interest and compound interest, over n periods. The annuity factor therefore corresponds to the reciprocal of the present value factor. (a) (annually) in arrears Aarrears =

qn · (q − 1) qn · i = qn − 1 qn − 1

(b) (annually) in advance Aadvance =

qn · i q · (qn − 1)

Example: Mrs. Penny will inherit $1,000,000 on January 1st , 2010. She would like to consume this amount in equal parts every year for the next 15 years. In doing so, she calculates with a calculation interest rate (average interest rate during these 15 years) of 2.5 % p.a. Mrs. Penny would like to have the yearly annuity paid out at the beginning of each (calendar) year. This annual, in advance annuity factor is calculated as follows Aadvance =

(1.025)15 · 0.025 qn · i = ≈ 0.078797 q · (qn − 1) (1.025) · (1.02515 − 1)

⇒ Radv. 15 ≈ $78, 797 Mrs. Penny has around $78,797 available at the beginning of each (calendar) year for 15 years, if she would like to consume the sum of $1,000,000 as planned at a calculation interest rate of 2.5 % p.a.

7.4 Annuity Calculation Overview:

7.4.2 Finite, Regular Annuity 7.4.2.1 Annual Annuity with Annual Interest Annuity and interest periods are exactly one year. (a) Annuity in arrears: Payment due at the end of the year

Amount of Annuity

R1 = r1

R2 = r2 + R1 · q = r2 + r1 · q R3 = r3 + R2 · q = r3 + r2 · q + r1 · q2 .. .

183

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184

Rn = rn + rn−1 · q + rn−2 · q2 + ... + r2 · qn−2 + r1 · qn−1 Rn = r · (1 + q1 + q2 + ... + qn−1 ) Rn = r ·

qn − 1 qn − 1 = r· q−1 i | {z }

final annuity value factor

Present Value of Annuity

Cn = C0 · qn = b Rn = R0 · qn

R0 = r ·

qn − 1 qn − 1 = r · qn · (q − 1) qn · i | {z }

annuity present value factor

Annuity (Rn given) r = Rn ·

q−1 i = Rn · n qn · (q − 1) q −1

Annuity (R0 given) r = R0 ·

i · qn (q − 1) · qn = R · 0 qn − 1 qn − 1

Interest Rate (Rn given)

Rn = r ·

qn − 1 i

Calculation of the zeros of the function f (i) = −Rn + r · f ′ (i) = r ·

(1 + i)n − 1 i

i · n · qn−1 − qn + 1 i2

Newton’s method of approximation ik+1 = ik −

f (ik ) f ′ (ik )

185

7.4 Annuity Calculation

Interest Rate (R0 given)

R0 = r ·

qn − 1 qn · (q − 1)

Calculation of the zeros of the function f (i) = −R0 + r · f ′ (i) = r ·

(1 + i)n − 1 i · (1 + i)n

q + n · i − qn+1 i2 · qn+1

Newton’s tangent method ik+1 = ik − Period (Rn given)

Rn = r ·

f (ik ) f ′ (ik )

qn − 1 i

qn = 1 +

i · Rn r

| ·i; ÷r; +1 | ln(...)

  i · Rn ln 1 + r n= ln(q) Period (R0 given)

qn − 1 i · qn   i · Rn 1 n = ln +1 · r ln(q)

R0 = r ·

(b) Annuity in advance: Payment due at the beginning of the year

Amount of Annuity

R1 = r0 · q

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186

R2 = r1 · q + R1 · q = r1 · q1 + r0 · q2 R3 = r2 · q + R2 · q = r2 · q1 + r1 · q2 + r0 · q3 .. . Rn = rn−1 · q1 + rn − 2 · q2 + ... + r2 · qn−2 + r1 · qn−1 +r0 · qn Rn = r · (q1 + q2 + ... + qn ) Rn = r · q · (1 + q1 + q2 + ... + qn−1 ) Rn = r · q · Present Value of Annuity

R0 = r ·

qn − 1 qn − 1 = r·q· q−1 i

q · (qn − 1) i · qn

Annuity (Rn given) r = Rn ·

q−1 Rn · i = q · (qn − 1) q · (qn − 1)

Annuity (R0 given) r = R0 ·

(q − 1) · qn i · qn = R0 · n q · (q − 1) q · (qn − 1)

Interest Rate (Rn given)

f (i) = −Rn + r · f ′ (i) = r ·

Interest Rate (R0 given)

(1 + i)n+1 − (1 + i) i

i · (n + 1) · qn − qn+1 + 1 i2

f (i) = −R0 + r ·

(1 + i)n+1 − (1 + i) i · (1 + i)n

187

7.4 Annuity Calculation i · ((n + 1) · n) − q + q1−n i2   i · Rn ln q + r n= −1 ln(q) f ′ (i) = r ·

Period (Rn given)

Period (R0 given)

  i · R0 ln q − r n = 1− ln(q)

7.4.2.2 Annual Annuity with Sub-Annual Interest The annuity periods cover one year (mr = 1), however there are multiple interest periods per year (mz > 1). Nominal, relative and conforming interest rate Conforming Interest Rate

Relative Interest Rate

∗

mz

discrete

i = (1 + j)

steady

i∗ = ei − 1

1

j = (1 + i∗ ) mz − 1

(a) Annuity in arrears:

Amount of Annuity

Rn = r ·

(1 + i∗ )n − 1 i∗

Present Value of Annuity

R0 = r ·

(1 + i∗ )n − 1 i∗ · (1 + i∗ )n

 −1 =

i 1+ mz

mz −1

7 Financial Mathematics

188

Annuity (Rn given)

r = Rn ·

i∗ (1 + i∗ )n − 1

Annuity (R0 given)

r = Rn ·

i∗ · (1 + i∗ )n (1 + i∗ )n − 1

Interest Rate (Rn given)

f (i∗ ) = −Rn + r · f ′ (i∗ ) = r ·

Interest Rate (R0 given)

i∗ · n · (1 + i∗ )n−1 − (1 + i∗ )n + 1 (i∗ )2

f (i∗ ) = −R0 + r · f ′ (i∗ ) = r ·

(1 + i∗ )n − 1 i∗

(1 + i∗ )n − 1 i∗ · (1 + i∗ )n

(1 + i∗ ) + n · i∗ − (1 + i∗ )n+1 (i∗ )2 · (1 + i∗ )n+1

 i∗ · Rn ln 1 + r n= ln(1 + i∗ ) ∗  1 i · Rn +1 · n = ln r ln(q) 

Period (Rn given) Period (R0 given)

Example: An annuity of $700 payable annually in arrears is paid over a period of eight years. The interest rate is 1.25 % per quarter. What is the present value? i∗ = (1 + 0.0125)4 − 1 =

R0 = $700 ·

  0.05 1+ − 1 ≈ 0.05095 4

(1 + 0.05095)8 − 1 = $4, 506.93 0.05095 · (1 + 0.05095)8

The present value is $4,506.93.

189

7.4 Annuity Calculation (b) Annuity in advance:

Amount of Annuity

Rn = r ·

Present Value of Annuity

R0 = r ·

Annuity (Rn given) r = Rn · Annuity (R0 given) r = R0 · Interest Rate (Rn given)

(1 + i∗ )n+1 − (1 + i∗ ) i∗ (1 + i∗ )n − 1 i∗ · (1 + i∗ )n−1 i∗ (1 + i∗ )n+1 − (1 + i∗ ) i∗ · (1 + i∗ )n−1 (1 + i∗ )n − 1

f (i∗ ) = −Rn + r ·

(1 + i∗ )n+1 − (1 + i∗ ) i∗ ∗ )n ·i∗ −i∗ −(1+i∗ )n+1 +(1+i∗ )

f ′ (i∗ ) = r · (n+1)·(1+i Interest Rate (R0 given)

f (i∗ ) = −R0 + r ·

f ′ (i∗ ) = r ·

Period (Rn given)

Period (R0 given)

(i∗ )2

(1 + i∗ )n − 1 i∗ · (1 + i∗ )n−1

n · i∗ − (1 + i∗ ) − i∗ · (n − 1) (i∗ )2

  i∗ · Rn ∗ ln (1 + i ) + r n= −1 ∗ ln(1 + i )   i∗ · R0 ln (1 + i∗ ) − r n = 1− + ∗ ln( i )

7 Financial Mathematics

190 Example:

An annual annuity of $700 payable in advance is paid over a period of eight years. The interest rate is 1.25 % per half year. What is the amount of annuity? i∗

  0.025 2 = (1 + 0.0125) − 1 = 1 + − 1 ≈ 0.02516 2 2

R8 = $700 ·

(1 + 0.02516)8+1 − (1 + 0.02516) = $4, 585.76 0.02516

The amount of annuity after eight years is $4,585.76.

7.4.2.3 Sub-Annual Annuity with Annual Interest Annuities are paid in sub-annual annuity periods (semi-annual mr = 2, quarterly mr = 4, monthly mr = 12), the interest periods cover one year. T

is equivalent to the amount of the regular annuity payments

(a) Integer periods Annuity payment in arrears  i i h T = r · mr + · 0 + 1 + 2 + ... + (mr − 1) mr with 0 + 1 + 2 + ... + (mr − 1) =

(mr − 1) · mr 2

   i (mr − 1) · mr T = r · mr + · mr 2

191

7.4 Annuity Calculation  i T = r · mr + · (mr − 1) 2 

Amount of Annuity

  n i q −1 Rn = r · mr + · (mr − 1) · 2 i | {z } T

Present Value of Annuity

 n  i q −1 R0 = r · mr + · (mr − 1) · 2 i · qn

Annuity payment in advance  i i h T = r · mr + · 1 + 2 + 3 + ... + mr mr with 1 + 2 + 3 + ... + mr =

(mr + 1) · mr 2

   i (mr + 1) · mr T = r · mr + · mr 2  i T = r · mr + · (mr + 1) 2 

Amount of Annuity

  n i q −1 Rn = r · mr + · (mr + 1) · 2 i | {z } T

Present Value of Annuity

  n i q −1 R0 = r · mr + · (mr + 1) · 2 i · qn

7 Financial Mathematics

192 Example:

Mrs. Penny inherits a monetary amount, which she invests at an interest rate of 4 % p.a. She would like to be paid a constant monthly amount of $1,400 in advance for ten years, so that the money is used up at the end of the period. How much money did she inherit?   0.04 1.0410 − 1 R0 = $1, 400 · 12 + · (12 + 1) · = $139, 215.42 2 0.04 · 1.0410 The amount of money that Mrs. Penny inherited is $139,215.42.

(b) Non-integer periods

n=

N mr

N∈Z

N1 = n1 · mr

n1 = int(n)

and

with

N2 = n2 · mr

n2 = n − n1

Annuity payment in arrears Amount of Annuity Rn = r ·

h  n  i 1 2 mr + 2i · (mr − 1) · q i−1 · (1 + n2 · i) + N2 + m1r · (N2 −1)·N 2

Present Value of Annuity   n   i q 1 −1 1 (N2 − 1) · N2 mr + · (mr − 1) · · (1 + n2 · i) + · 2 i mr 2 R0 = r · qn1 · (1 + n2 · i)

7.4 Annuity Calculation

193

Annuity payment in advance Amount of Annuity Rn = r ·

h  n  i 1 2 mr + 2i · (mr + 1) · q i−1 · (1 + n2 · i) + N2 + m1r · (N2 +1)·N 2

Present Value of Annuity   n   q 1 −1 1 (N2 + 1) · N2 i · (1 + n2 · i) + · mr + · (mr + 1) · 2 i m2 2 R0 = r · n 1 q · (1 + n2 · i) Example: Mrs. Penny deposits an amount of $500 each quarter in arrears into an account. The interest rate is 2 % p.a. How much money is in the account after 10 years and 6 months? N2 = 0.5 · 4 = 2   1.0210 − 1 0.02 12 + · (12 − 1) · · (1 + 0.5 · 0.02) 2 0.02   1 (2 − 1) · 2 + 2+ · 12 2

R10.5 = $500 ·

R10.5 = $68, 005.23 After 10 years and 6 months there are $68,005.23 in the account.

7 Financial Mathematics

194

7.4.2.4 Sub-Annual Annuity with Sub-Annual Interest Annuity and interest periods are shorter than one year. (a) Annuity period = interest period Annuity payment in arrears

Amount of Annuity

RN = r ·

(1 + j)N − 1 qN − 1 qN − 1 = r· = r· j j j

Present Value of Annuity

R0 = r ·

qN − 1 (1 + j)N − 1 qN − 1 = r · = r · qN · (q − 1) (1 + j)N · j j · qN

Annuity payment in advance

Amount of Annuity

RN = r · q ·

Present Value of Annuity

R0 = r ·

with j =

i mz

qN − 1 j

q · (qN − 1) j · qN

sub-annual interest rate

N = mr · n period of annuity in the sub-annual periods n = number of relevant years mz = number of sub-annual interest periods mr = number of sub-annual annuity periods here mr = mz , since annuity periods = interest rates

7.4 Annuity Calculation

195

Example 1: Camilla saves an amount of $100 at the end of each month, which earns interest at 1.2% p.a. per month. What is her capital after 1.5 years? RN = ? with r = $100 j =

⇒ RN = 100 ·

0.012 = 0.001 N = 12 · 1.5 = 18 q = 1.001 12

(1.001)18 − 1 = $1, 815.38 0.001

Example 2: Michelle inherits $20,000, which earns interest at her commercial bank ˙ p.a. per month. She wants to withdraw a constant amount at the at 3% beginning of each month for two years. How much is this monthly annuity payment in advance? r= ? R0 = r ·

q · (qN − 1) j · qN

⇒ r = R0 ·

j · qN q · (qN − 1)

with R0 = $20, 000 j=

0.03 = 0.0025 12

q = 1.0025

7 Financial Mathematics

196 N = 12 · 2 = 24

0.0025 · 1.002524 = $857.48 1.0025 · (1.002524 − 1)

r = 20, 000 ·

Example 3: Steven wants to deposit an amount of $500 into his account at the be˙ each month. ginning of each month, which will earn interest at 0.25% His goal is to have $10,000 at the end of the period of this annuity. How long does Steven have to pay in? n= ? RN = r ·

q · (qN − 1) (in advance) j

with N = mr · n ⇒ RN = r ·

q · (qmr ·n − 1) j

⇔

RN · j = q · (qmr ·n − 1) r

⇔

RN · j + 1 = qmr ·n r·q 

RN · j +1 ⇔ ln r·q 

 = mr · n · lnq

 RN · j ln +1 r·q ⇔ n= lnq · mr

197

7.4 Annuity Calculation with RN = $10, 000 r = $500 mz = 0.0025 bzw. j = 0.0025 · 12 = 0.03 = 3 % p.a. q = 1.0025 N = mr · n with mr = 12 N = period in months n = period in years

 ln n=

 10, 000 · 0.0025 +1 500 · 1.0025 ≈ 1.6244 Jahre ln1.0025 · 12

N = 12 · 1.6244 ≈ 19.4929 months

(b) Annuity period < interest period Annuity payment in arrears N  ∗ 1 + mi r −1 (1 + j∗ )N − 1 RN = r · = r· ∗ i ∗ j m

Amount of Annuity

with

r

i∗ = conforming interest rate j∗ =

i∗ mr

7 Financial Mathematics

198

  i mz −1 1+ mz

discrete

i∗ = (1 + j)mz − 1 =

continuous

i∗ = e i − 1

and

j = relative interest rate j=

1 mz

or

j∗ = (1 + j)mz /mr − 1

and

N = mr · n

Example: Nawid deposits an amount of $300 into a savings account at the end ˙ per month. What is his of each quarter, which earns interest at 0.25% capital after 1.5 years?

RN = ? with r = $300 mr = 4 mz = 12 n = 1.5 N = mr · n = 4 · 1.5 = 6 j = 0.0025 j∗ = (1 + j)mz /mr − 1 = = (1.0025)12/4 − 1 = 0.007519 RN = r ·

(1 + j∗ )N − 1 j∗

199

7.4 Annuity Calculation R6 = 300 ·

(1.007519)6 − 1 = $1, 834.18 0.007519

The following table illustrates the development of the capital for the example above. Quarter

Capital at the beginning of the quarter

Interest

Annuity Capital at the end of the quarter

1

0.00

0.00

300

300.00

2

300.00

300.00 ·

300

602.26

300

906.79

300

1, 213.61

300

1, 522.74

300

1, 834.1911

· 0.007519 = 2.26 3

602.26

602.26 · · 0.007519 = 4.53

4

906.79

906.79 · · 0.007519 = 6.82

5

1, 213.61

1, 213.61 · · 0.007519 = 9.13

6

1, 522.74

1, 522.74 · · 0.007519 = 11.45

11

Rounding error above 1 cent.

7 Financial Mathematics

200

 N i∗ 1 + −1 mr (1 + −1 R0 = r · = r·  N N j∗ · (1 + j∗ ) i∗ i∗ mr · 1 + mr j ∗ )N

Present Value of Annuity

j∗ =

with

i∗ mr

Example: ˙ per month at Eva wins an amount of money that earns interest at 0.2% her commercial bank. She wants to be paid $6,000 at the end of each half-year for two years, so that this amount of money will be used up at the end of the 4th semester, i.e., after two years. What is the amount of money she has won?

R0 = ? with r = $6, 000 mr = 2 mz = 12 n = 2 N = mr · n = 2 · 2 = 4

j = 0.002

j∗ = (1 + j)mz /mr − 1 = = (1.002)12/2 − 1 = 0.01206 R0 = r ·

(1 + j∗ )N − 1 j∗ · (1 + j∗ )N

= 6, 000 ·

(1.01206)4 − 1 0.01206 · (1.01206)4

= $23, 293.49

201

7.4 Annuity Calculation

The following table illustrates the development of the capital for the example above.

Half-year

Capital at the beginning of the half-year

Interest

Annuity

1

23, 293.49

23, 293.49 ·

6, 000

0.01206 =

Capital at the end of the half-year 23, 293.49 +280.92 −6, 000.00

280.92

= 17, 574.41 2

17, 574.41

17, 574.41 ·

6, 000

17, 574.41

0.01206 =

+211.95

= 211.95

−6, 000.00 = 11, 786.36

3

11, 786.36

11, 786.36 ·

6, 000

0.01206 =

11, 786.36 +142.14 −6, 000.00

142.14

= 5, 928.50 4

5, 928.50

5, 928.50

6, 000

5, 928.50

· 0.01206

+71.50

= 71.50

−6, 000.00 = 0.00

7 Financial Mathematics

202 Annuity payment in advance RN = r · (1 + j∗ ) ·

Amount of Annuity

= r·

(1 + j∗ )N+1 − (1 + j∗ ) j∗ 

RN = r · 1 +

or

 = r·

j∗ =

with

(1 + j∗ )N − 1 j∗

i∗



mr ∗

1 + mi r

·

N  ∗ −1 1 + mi r i∗ mr

N+1

  ∗ − 1 + mi r

i∗ mr

i∗ mr

Example: At the beginning of each quarter, Paul deposits an amount of $300 into a savings account, which earns interest at a rate of 0.25 % per month. What is his capital after 1.5 years? RN = ?

with r = $300 mr = 4 mz = 12 n = 1.5 N = mr · n = 4 · 1.5 = 6 j∗ = (1 + j)mz /mr − 1 =

j = 0.0025

203

7.4 Annuity Calculation = (1.0025)12/4 − 1 = 0.007519

RN = r ·

(1 + j∗ )N+1 − (1 + j)∗ j∗

R6 = 300 ·

(1.007519)6+1 − (1.007519) 0.007519

= $1, 847.97 = R6in arrears · 1.007519 = $1, 834.18 · 1.007519

Present Value of Annuity

R0 = r · (1 + j∗ ) ·

= r·

= r·

or

(1 + j∗ )N − 1 j∗ · (1 + j∗ )N

(1 + j∗ )N+1 − (1 + j∗ ) j∗ · (1 + j∗ )N

=

=

(1 + j∗ )N − 1 j∗ · (1 + j∗ )N−1

    1 + i∗ N − 1 mr i∗ R0 = r · 1 + ·  N = mr i∗ i∗ · 1 + mr mr  = r·

∗

1 + mi r i∗ mr

N+1

  ∗ − 1 + mi r =  N ∗ · 1 + mi r

7 Financial Mathematics

204 N ∗ −1 1 + mi r = r· N−1  i∗ i∗ mr · 1 + mr 

j∗ =

with

i∗ mr

Example: Maria wins an amount of money that earns her interest at her commercial bank at 0.2 % per month. She wants to be paid $6,000 at the beginning of each half-year for two years, so this amount of money will be used up by the beginning of the fourth semester. What is the amount of money she has won? R0 = ? with r = $6, 000 mr = 2 mz = 12 n = 2 N = mr · n = 2 · 2 = 4

j = 0.002

j∗ = (1 + j)mz /mr − 1 = (1.002)12/2 − 1 = 0.01206 R0 = r ·

(1 + j∗ )N − 1 j∗ · (1 + j∗ )N−1

= 6, 000 ·

=

(1.01206)4 − 1 0.01206 · (1.01206)3

= $23, 574.41

arrears · 1.01206 = $23, 293.49 · 1.01206 = Rin 0

(c) Annuity period > interest period Annuity payment in arrears

7.4 Annuity Calculation   j (1 + j)N − 1 RN = r · mr + · (mr − 1) · 2 j

Amount of Annuity

with

205

j = relative interest rate j=

1 mz

N = mz · n

and

mz = number of sub-annual interest periods = = number of interest periods p.a. mr = number of annuity periods per interest periods

Example: Nawid deposits an amount of $300 into a savings account at the end of each month, which earns interest at 0.25 % per quarter. What is his capital after 1.5 years? RN = ? with r = $300 mr = number of annuity periods per interest periods = 3 mz = number of interest periods p.a. = 4 N = mz · n = 4 · 1.5 = 6 mit n = 1.5

7 Financial Mathematics

206 j = 0.0025

  j (1 + j)N − 1 RN = r · mr + · (mr − 1) · 2 j   1.00256 − 1 0.0025 R6 = 300 · 3 + · (3 − 1) · = $5, 438.39 2 0.0025

Alternative Calculation Using the ICMA Method In financial practice, it may be required by law to use a specific method for a mixed interest rate (see chapter 7.1.2.3). As an example, the ICMA method will be followed in the following.12 The relative interest rate j is then to be adjusted to the annuity period as follows: (1 + j∗ )mr = (1 + j)3 = 1.0025 with j∗ = the relative interest rate emancipated over mr · mz · n periods ⇒ q∗ = (1 + j∗ ) = 1.00251/3 = 1.00083264 ⇒ j∗ = 0.00083264 = 0.83264 ‰ monthly RN = r ·

(q∗ )mr · mz · n − 1 j∗

R6 = 300 ·

12

(1.00083264)3 · 4 · 1.5 − 1 = $5, 438.39 0.00083264

For Germany according to §6 PAngV.

7.4 Annuity Calculation

Present Value of Annuity

207

  j (1 + j)N − 1 R0 = r · mr + · (mr − 1) · 2 j · (1 + j)N

with j = relative interest rate j=

1 mz

and N = mz · n

mz = number of sub-annual interest periods = = number of interest periods p.a. mr = number of annuity periods per interest periods

Example: Eva wins an amount of money that earns interest at her commercial ˙ per half-year, i.e. per semester. She would like to be paid bank at 1.2% out $50 at the end of each month for two years, so that this amount of money will be used up at the end of the 4th semester, i.e., after two years. What is the amount of money she has won?

R0 = ? with r = $50 mr = number of annuity periods per interest periods = 6 mz = number of interest periods p.a. = 2 N = mz · n = 2 · 2 = 4

7 Financial Mathematics

208 with n = 2 j = 0.012  (1 + j)N − 1 j R0 = r · mr + · (mr − 1) · = 2 j · (1 + j)N 

 1.0124 − 1 0.012 · (6 − 1) · = $1, 170.67 = 50 · 6 + 2 0.012 · 1.0124 

Alternative Calculation Using the ICMA Method Cf. chapter 7.1.2.3 The relative interest rate j is then to be adjusted to the annuity period as follows: (1 + j∗ )mr = (1 + j)6 = 1.012 with j∗ = the relative interest rate emancipated over mr · mz · n periods ⇒ q∗ = (1 + j∗ ) = 1.0121/6 = 1.001990073 ⇒ j∗ = 0.001990073 = 1.990073 ‰ monthly R0 = r ·

(q∗ )mr · mz · n − 1 j∗ · (q∗ )mr · mz · n

R0 = 50 ·

13

(1.001990073)6 · 2 · 2 − 1 0.001990073 · (1.001990073)6 · 2 · 2

= 1, 170.66 13

The difference from the original calculation using the formula   j (1 + j)N − 1 R0 = r · mr + · (mr − 1) · is 1 cent. 2 j · (1 + j)N

7.4 Annuity Calculation

209

Annuity payment in advance

Amount of Annuity

  j (1 + j)N − 1 RN = r · mr + · (mr + 1) · 2 j

with j = relative interest rate j=

1 mz

and N = mz · n

mz = number of sub-annual interest periods = = number of interest periods p.a. mr = number of annuity periods per interest periods

Example: Paul deposits an amount of $100 into a savings account at the beginning of each month, which earns interest at 0.75 % each quarter. What is his capital after 1.5 years? RN = ?

with r = $100 mr = number of annuity periods per interest periods = 3 mz = number of interest periods p.a. = 4

7 Financial Mathematics

210 N = mz · n = 4 · 1.5 = 6 with n = 1.5 j = 0.0075

  j (1 + j)N − 1 RN = r · mr + · (mr + 1) · 2 j   1.00756 − 1 0.0075 R6 = 100 · 3 + · (3 + 1) · = $1, 843.26 2 0.0075

Alternative Calculation Using the ICMA Method Cf. chapter 7.1.2.3 The relative interest rate j is then to be adjusted to the annuity period as follows: (1 + j∗ )mr = (1 + j)3 = 1.0075 with j∗ = the relative interest rate emancipated over mr · mz · n periods ⇒ q∗ = (1 + j∗ ) = 1.00751/3 = 1.002493776 ⇒ j∗ = 0.002493776 = 2, 493776 ‰ monthly RN = r ·

q∗ · [(q∗ )mr · mz · n − 1] j∗

211

7.4 Annuity Calculation h i 1.002493776 · (1.002493776)3 · 4 · 1.5 − 1 R6 = 100 ·

0.002493776

Present Value of Annuity

= $1, 843.25 14

  (1 + j)N − 1 j R0 = r · mr + · (mr + 1) · 2 j · (1 + j)N

with j = relative interest rate j=

1 mz

and N = mz · n mz = number of sub-annual interest periods= = number of interest periods p.a. mr = number of annuity periods per interest periods

Example: Maria wins an amount of money that earns her interest at her commercial bank at 0.6 % quarterly. She wants to be paid $50 at the beginning of each month for two years, so that this amount of money will be used up at the beginning of the 24th month. What is the amount of money she has won?

14

The difference from the original calculation using the formula   j (1 + j)N − 1 RN = r · mr + · (mr + 1) · is 1 cent. 2 j

7 Financial Mathematics

212 R0 = ? with r = $50

mr = number of annuity periods per interest periods = 3 mz = number of interest periods p.a. = 4 N = mz · n = 4 · 2 = 8 mit n = 2 j = 0.006   (1 + j)N − 1 j R0 = r · mr + · (mr + 1) · = 2 j · (1 + j)N  1.0068 − 1 0.006 · (3 + 1) · = $1, 172.91 = 50 · 3 + 2 0.006 · 1.0068 

Alternative Calculation Using the ICMA Method Cf. chapter 7.1.2.3 The relative interest rate j is then to be adjusted to the annuity period as follows: (1 + j∗ )mr = (1 + j)3 = 1.006 with j∗ = the relative interest rate emancipated over mr · mz · n periods ⇒ q∗ = (1 + j∗ ) = 1.0061/3 = 1.001996013 ⇒ j∗ = 0.001996013 = 1.996013 ‰ monthly

213

7.4 Annuity Calculation R0 = r ·

q∗ · [(q∗ )mr · mz · n − 1] j∗ · (q∗ )mr · mz · n h i 1.001996013 · (1.001996013)3 · 4 · 2 − 1

R0 = 50 ·

= $1, 172.91

0.001996013 · (1.001996013)3 · 4 · 2

7.4.3 Finite, Variable Annuity These are recurring payments which are made at regular intervals and the extent of which changes over the course of time. The amount of the annuity changes over time. Changes of annuity can • vary without system (irregular annuity) • be systematic (regular annuity) – arithmetic progressive annuity – geometric progressive annuity

7.4.3.1 Irregular Annuity n

Amount of Annuity

Rn = qn ·

∑ rk · q−k

k=1 n

∑ rk · q−k

Present Value of Annuity

R0 =

Interest Rate (Rn given)

f (i) = −Rn + qn ·

k=1

n

∑ rk · q−k

k=1

f ′ (i) = qn−1 ·

n

∑ (n − k) · rk · q−k

k=1

7 Financial Mathematics

214 n

Interest Rate (R0 given)

f (i) = −R0 +

∑ rk · q−k

k=1 n

f ′ (i) = −q−1 ·

∑ k · rk · q−k

k=1

Example 1: Camilla pays irregular amounts into her savings account at the beginning of each year for three years: 1th year:

$1, 000

2nd

year:

$2, 500

3rd

year:

$3, 200

The credit balance earns interest at 2 % p.a. What amount of money does Camilla have after three years? r1 = $1, 000 r2 = $2, 500 r3 = $3, 200 n = 3 years q = 1.02 k = 3 R0 = $1, 000 · 1.02−1 + $2, 500 · 1.02−2 + $3, 200 · 1.02−3 = $6, 398.74 Rn = R3 = 6, 398.74 · 1.023 = $6, 790.39 After three years, Camilla has $6,790.39.

215

7.4 Annuity Calculation Example 2:

The conditions from example 1 (see above) apply. That means Camilla pays the following amounts at the beginning of each year over three years: r1 = $1, 000 r2 = $2, 500 r3 = $3, 200 n = 3 years R0 = $6, 398.74 Rn = R3 = $6, 790.39 The interest rate i is not known and is to be determined approximately using Newton’s method of approximation. The market suggests that this interest rate is likely to be between 1 % and 4 %. Therefore, i = 0.03 is to be assumed as a fictitious initial value.

Newton’s Method of approximation15 is to be applied with ik+1 = ik −

f (ik ) f ′ (ik )

ik

corresponds to the (fictitious) initial value tiously assumed) approximate value

ik+1

corresponds to the next (calculated) approximate value

15

See chapter 4.5.2.

=

first (ficti-

7 Financial Mathematics

216 a) at a given amount of annuity Rn f (i) = −Rn + r ·

f ′ (i)

 n  q −1 qn − 1 d + · −n i i i

    r qn − 1 d qn − 1 n−1 n−1 = · nq − + 2 · n + nq −2 i i i i

b) at a given present value of annuity R0  n  qn − 1 d q −1 −n f (i) = −R0 + r · + · −n·q i · qn i i · qn f ′ (i) =

ik = 0.03

    n qn − 1 d n qn − 1 r · n+1 − + · (1 + q + in) − 2 i q iqn i2 qn+1 iqn

fictitiously assumed initial value within the interval [1 %; 4 %]

ik+1 = ?

1. Deterine approximate value i = 0.03 (estimated) q = 1.03 Rn = $6, 790.39

determine: f (0.03)

determine: f ′ (0.03)

217

7.4 Annuity Calculation n

n

∑ rk · q−k

∑ (n − k) · rk · q−k

k=1

k=1

k = 1 : 1, 000 · 1.03−1

= 970.874

k = 1 : (3−1)·1, 000·1.03−1 = 1, 941.75

k = 2 : 2, 500 · 1.03−2

= 2, 356.49

k = 2 : (3−2)·2, 500·1.03−2 = 2, 356.49

k = 3 : 3, 200 · 1.03−3

= 2, 928.45

k = 3 : (3−3)·3, 200·1.03−3 = 0

∑ 6, 255.81

∑ 4, 298.24

1.033 · 6, 255.81 = 6, 835.89

f (0.03) = −6, 790.39 + 6, 835.89 =

f ′ (0.03) = 1.033−1 · 4, 298.24 =

= 45.5

in+1 = 0.03 −

= 4, 560

45.5 = 0.020022 4, 560

i1 = 0.020022 = first approximate value = new initial value

2. Determine approximate value i1 = 0.020022 q = 1.020022 Rn = 6, 790.39

determine: f (0.020022)

n

∑ rk · q−k k=1

determine: f ′ (0.020022)

n

∑ (n − k) · rk · q−k k=1

7 Financial Mathematics

218 k = 1 : 1, 000 · 1.020022−1 = 980.371

k = 1 : (3 − 1) · 1, 000 · 1.020022−1 = 1, 960.74

k = 2 : 2, 500 · 1.020022−2 = 2, 402.82

k = 2 : (3 − 2) · 2, 500 · 1.020022−2 = 2, 402.82

k = 3 : 3, 200 · 1.020022−3 = 3, 015.24

k = 3 : (3 − 3) · 3, 200 · 1.020022−3 = 0

∑ 6, 398.43

∑ 4, 363.56

1.0200223 · 6, 398.43 = 6, 790.5

f (0.020022) = −6, 790.39 + 6, 790.5 =

f ′ (0.020022) = 1.0200223−1 · 4, 363.56 =

= 0.11

in+1 = 0.020022 −

= 4, 540.04

0.11 = 0.019998 4, 540.04

i2 = 0.019998

3. Determine approximate value i2 = 0.019998 q = 1.019998 Rn = 6, 790.39

determine: f (0.019998)

n

∑ rk · q−k k=1

k = 1 : 1, 000 · 1.019998−1 = 980.384

determine: f ′ (0.019998)

n

∑ (n − k) · rk · q−k k=1

k = 1 : (3 − 1) · 1, 000 · 1.019998−1 = 1, 960.79

219

7.4 Annuity Calculation k = 2 : 2, 500 · 1.019998−2 = 2, 402.93

k = 2 : (3 − 2) · 2, 500 · 1.019998−2 = 2, 402.93

k = 3 : 3, 200 · 1.019998−3 = 3, 015.45

k = 3 : (3 − 3) · 3, 200 · 1.019998−3 = 0

∑ 6, 398.77

∑ 4, 363.72

1.0199983 · 6, 398.77 = 6, 790.39

f (0.019998) = −6, 790.39 + 6, 790.39 =

f ′ (0.019998) = 1.0199983−1 · 4, 363.72 =

=0

in+1 = 0.019998 −

= 4, 540

0 = 0.019998 4, 540

i3 = 0.019998

Solution By (iteratively) continuing the determination of further approximate values, the following is obtained: i1 = 0.020022 i2 = 0.019998 (≈ 0.02 → rounded to two decimal places) i3 = 0.019998 (≈ 0, 02 → rounded to two decimal places) → The result for the interest rate i is 0.019998, i.e. approximately i = 0.02 or p = 2 %

7 Financial Mathematics

220 7.4.3.2 Arithmetic Progressive Annuity

The annuity payment increases from period to period by a predetermined amount.

r

equivalent to the annuity payment at point in time k = 1

d

equivalent to the difference between two successive annuity payments d = rk+1 − rk

Amount of Annuity

r1 = r r2 = r + d r3 = r + 2d .. . rn = r + (n − 1) · d n

Rn =

∑

h

i r + (k − 1) · d · qn−k

k=1

 n  qn − 1 d q −1 Rn = r · + · −n i i i

Present Value of Annuity

R0 = r ·

Annuity (Rn given)

r=

 n  qn − 1 d q −1 −n + · − n · q i · qn i i · qn

Rn · i + d · n d − qn − 1 i

221

7.4 Annuity Calculation

Annuity (R0 given)

Interest Rate (Rn given)

r=

R0 · i · qn + d · n d − qn − 1 i

 n  qn − 1 d q −1 f (i) = −Rn + r · + · −n i i i

Interest Rate (R0 given)

f (i) = −R0 + r ·

 n  qn − 1 d q −1 −n + · − n · q i · qn i i · qn

Period (Rn given)

f (n) = −Rn + r ·

 n  qn − 1 d q −1 + · −n i i i

Period (R0 given)

f (n) = −R0 + r ·

 n  qn − 1 d q −1 −n + · − n · q i · qn i i · qn

Example 1: Mrs. Penny would like to save money for her newborn godchild, Camilla, which she will receive on her 18th birthday. She starts to deposit $500 in the first year, but would like to increase the amount by $50 each year. The interest rate is 0.7 % p.a. How much money will Camilla receive on her 18th birthday? Rn = $500 ·

  1.00718 − 1 $50 1.00718 − 1 + · − 18 0.007 0.007 0.007

Rn ≈ $17, 499.27 Camilla is entitled to $17,499.27 at the end of the 18 years, if Mrs. Penny - as intended - deposits the increasing amount into the account each year.

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222 Example 2:

Steven receives an annuity (= annuity payment) over 8 years. The first annuity payment is $2,000, which is increased by $100 per year. The interest rate is 5 % p.a. What is the present value of this annuity? n = 8 i = 0.05 q = 1.05 r = $2, 000 d = $100 R0 = 2, 000 ·

  1.058 − 1 100 1.058 − 1 −8 + · − 8 · 1.05 0.05 · 1.058 0.05 0.05 · 1.058

= $15, 023.42 The present value of this arithmetic progressed annuity is $15,023.42.

Example 3: Mrs. Pfennig saved a total of $9,736.46 (Rn ) for her grandson for 18 years. The money earned interest at 0.5 % p.a. and she increased the amount by $50 annually. How much was the first annuity payment of this annuity?

Rn = $9, 736.46 i = 0.005 d = $50 n = 18 q = 1.005 r=

9, 736.46 · 0.005 + 50 · 18 50 − 1.00518 − 1 0.005

r = $100 The first annuity payment was $100.

7.4 Annuity Calculation

223

Example 4: Given the present value of annuity of $15,023.42, the annuity payment was increased by $100 per year for eight years at an interest rate of 5 % p.a. How much was the first annuity payment of this annuity?

R0 = $15, 023.42 i = 0.05 q = 1.05 d = $100 n = 8 r=

15, 023.42 · 0.05 · 1.058 + 100 · 8 100 − 1.058 − 1 0.05

r = 1, 999.9998 ≈ $2, 000 The first annuity payment was $2,000.

Calculation of the Interest Rate i Newton’s Method of approximation (see chapter 4.5.2) is to applied with ik+1 = ik −

f (ik ) f ′ (ik )

ik

corresponds to the (fictitious) initial value =first (fictitiously assumed) approximate value

ik+1

corresponds to the next (calculated) approximate value

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a) at a given amount of annuity Rn

 n  q −1 qn − 1 d + · −n f (i) = −Rn + r · i i i

f ′ (i) =

    r qn − 1 d qn − 1 · nqn−1 − + 2 · n + nqn−1 − 2 i i i i

b) at a given present value of annuity R0

 n  q −1 qn − 1 d −n + · −n·q f (i) = −R0 + r · i · qn i i · qn

f ′ (i)

    r n qn − 1 d n qn − 1 = · n+1 − + 2 · n+1 (1 + q + in) − 2 i q iqn i q iqn

Calculation of the Period n Newton’s Method of approximation (see chapter 4.5.2) is to applied with nk+1 = nk −

f (nk ) f ′ (nk )

7.4 Annuity Calculation

225

nk

corresponds to the (fictitious) initial value =first (fictitiously assumed) approximate value

nk+1

corresponds to the next (calculated) approximate value

a) at a given amount of annuity Rn

f (n) = −Rn + r ·

 n  q −1 qn − 1 d + · −n i i i

  d d qn lnq · r+ f ′ (n) = − + i i i

b) at a given present value of annuity R0

 n  qn − 1 d q −1 −n f (n) = −R0 + r · + · −n·q i · qn i i · qn

f ′ (n)

=

−q−n



  d lnq d − · r + nd + i i i

Example 1: Mrs. Pfennig wants to save money for her just-born godchild, Penny, which she will receive on her 18th birthday. She starts paying in $500 in the first year, but would like to increase the amount by $50 each year.

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The interest rate is 0.7% p.a. On her 18th birthday, i.e., after 18 years of life, Penny should have Rn = $17,499.27 (see example above) at her disposal. r = $500 d = $50 i = 0.007 (0.7%)

q = 1.007

As a fictitious (estimated) initial value n0 = 20 are to be assumed with qn = 1.00720 ≈ 1.1497

nk+1 = nk −

f (n) f ′ (n)

 n  qn − 1 d q −1 f (n) = −Rn + r · + · −n i i i   1.0020 − 1 50 1.00720 − 1 f (20) = −17, 499.27 + 500 · + · − 20 0.007 0.007 0.007 f (20) = 3, 105.65

f ′ (n)

  d qn · ln(q) d =− + · r+ i i i

f ′ (20)

  1.00720 · ln(1.007) 50 50 =− + · 500 + 0.007 0.007 0.007

f ′ (20) = 1, 613.62

nk+1 = nk − n0 = 20

f (x) f ′ (x)

7.4 Annuity Calculation n1 = 20 −

227

3, 105.65 1, 613.62

n1 = 18.075352 = first approximate value = new initial value

The approximate value n1 = 18.075352 is now inserted in place of the initially fictitiously assumed initial value of n0 = 20 into the following formula nk+1 = nk −

f (nk ) f ′ (nk )

This iterative procedure is continued until the approximate values change only marginally and converge to a fixed value. In the example above, the following values are obtained: n2 = 18.000114 n3 = 18.000000 n4 = 18.000000 Thus, the result is n = 18 years.

Example 2: Michelle deposited money into her account each month. In the first month she deposited $50, in the following months she increased her deposit by $25 each. The balance earned exponential interest at 1 % per month. Now, at the end of the period of this savings plan, Michelle gets Rn =$1,678.64 paid out. Over how many months n did this savings plan run? The exact period n is no longer known to Michelle and is to be determined approximately using Newton’s method of approximation. Michelle vaguely remembers that the period of this savings plan was between 8 and 13 months. Therefore, one year, i.e. 12 months, is to be assumed as a fictitious initial value.

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nk+1 = nk −

f (n) f ′ (n)

f (n) = −Rn + r ·

 n  q −1 qn − 1 d + · −n i i i

f (12) = −1, 678.64 + 50 ·

  1.0112 − 1 25 1.0112 − 1 + · − 12 = 0.01 0.01 0.01

= $661.734 f ′ (n)

  d qn · ln(q) d =− + · r+ i i i

f ′ (12)

  25 1.0112 · ln(1.01) 25 =− + · 50 + = 0.01 0.01 0.01 = $359.132

nk+1 = nk −

f (x) f ′ (x)

n0 = 12 n1 = 12 −

661.734 359.132

n1 = 10.1574 = first approximate value = new initial value

The approximate value n1 = 10.1574 is now inserted in place of the initially fictitiously assumed initial value of n0 = 12 into the following formula nk+1 = nk −

f (nk ) f ′ (nk )

229

7.4 Annuity Calculation

This iterative procedure is continued until the approximate values change only marginally and converge to a fixed value. In the example above, the following calculations and values are obtained:

2. Determine approximate value f (10.1574) = −1, 678.64 + 50 ·

1.0110.1574 − 1 25 + · 0.01 0.01

1.0110.1574 − 1 · − 10.1574 0.01 

f ′ (10.1574)

 = 48.0077

  25 1.0110.1574 · ln1.01 25 =− + · 50 + = 307.185 0.01 0.01 0.01

nk+1 = 10.1574 −

48.0077 307.185

n2 = 10.0011

3. Determine approximate value f (10.0011) = −1, 678.64 + 50 ·  ·

f ′ (10.0011) = −

1.0110.0011 − 1 25 + · 0.01 0.01

1.0110.0011 − 1 − 10.0011 0.01

 = 0.335072

  25 1.0110.0011 · ln1.01 25 + · 50 + = 302.826 0.01 0.01 0.01

nk+1 = 10.0011 −

0.335072 302.826

n3 = 9.99999 ≈ 10

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4. Determine approximate value f (9.99999) = −1, 678.64 + 50 ·  ·

f ′ (9.99999) = −

25 1.019.99999 − 1 + · 0.01 0.01

1.019.99999 − 1 − 9.99999 0.01

 = −0.001048

  25 1.019.99999 · ln1.01 25 + · 50 + = 302.795 0.01 0.01 0.01

nk+1 = 9.99999 −

−0.001048 302.795

n4 = 9.99999 ≈ 10

5. Determine approximate value (test) f (10) = −1, 678.64 + 50 ·

  25 1.0110 − 1 1.0110 − 1 + · − 10 = 0.01 0.01 0.01

= 0.00198 f ′ (10)

  25 1.0110 · ln1.01 25 =− + · 50 + = 302.796 0.01 0.01 0.01

nk+1 = 10 −

0.00198 302.796

n5 = 9.99999 ≈ 10

By continuing the approximation the following is obtained n1 = 10.1574 n2 = 10.0011

231

7.4 Annuity Calculation n3 = 9.99999 ≈ 10 n4 = 9.99999 ≈ 10 n5 = 9.99999 ≈ 10 The period n of this arithmetic progressive annuity was 10 months.

7.4.3.3 Geometric Progressive Annuity The annuity payment increases annually by a given percentage. g

is equivalent to the growth factor of two successive annuity payments g=

rk+1 rk

(a) Interest factor ̸= growth factor (q ̸= g)

Amount of Annuity

Rn = r ·

qn − gn q−g

Present Value of Annuity

R0 = r ·

qn − gn (q − g) · qn

Annuity (Rn given)

r = Rn ·

q−g qn − gn

Annuity (R0 given)

r = R0 ·

(q − g) · qn qn − gn

Interest Rate (Rn given)

f (i) = −Rn + r ·

qn − gn q−g

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Interest Rate (R0 given)

f (i) = −R0 + r ·

qn − gn (q − g) · qn

Period (Rn given)

f (n) = −Rn + r ·

qn − gn q−g

Period (R0 given)

f (n) = −R0 + r ·

qn − gn (q − g) · qn

Example: Mrs. Penny would like to know how the amount for her newborn godchild will change if she increases the first amount of $500 by 10 % in the following years. g=

$550 = 1.1 $500

Rn = $500 ·

1.00718 − 1.118 1.007 − 1.1

Rn ≈ $23, 796.41 If Mrs. Penny increases the annual amount by 10 % each year, Camilla would receive $23,796.41 on her 18th birthday.

(b) Interest factor = growth factor (q = g)

Amount of Annuity

Rn = r · n · qn−1

Present Value of Annuity

R0 =

r·n q

233

7.4 Annuity Calculation

Annuity (Rn given)

r=

Annuity (R0 given)

r=

Interest Rate (Rn given)

Rn n · qn−1

R0 · q n r n−1 Rn i= −1 r·n r·n −1 R0

Interest Rate (R0 given)

i=

Period (Rn given)

f (n) = −Rn + r · n · qn−1

Period (R0 given)

n=

q · R0 r

Example: Mrs. Penny wants to save $500 a year for Camilla again. This time, she would like to increase each annual savings amount by 5 %. She is budgeting at a calculatory interest rate of 5 % p.a. Rn = $500 · 18 · 1.118−1 Rn ≈ $45, 490.23 Camilla would receive $45.490,23 on her 18th birthday.

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7.4.4 Perpetuity A perpetuity is an annuity that can be generated from the interest income of a fixed-interest investment without reducing the amount of the invested capital. Accordingly, a perpetual annuity is characterized by an infinitely long flow of payments. Consequently, the amount of annuity becomes infinitely large. Since the capital is fully preserved, the resulting return is generated “eternally”. (a) Annuity in arrears

Present Value of Annuity

R0 = r ·

qn − 1 i · qn

with i > 0

 n  r q −1 R0 = · i qn  n  q 1 r R0 = · n − n i q q   r 1 R0 = · 1 − n i q if q > 1 and i > 0 :   1 r r · 1− n = n→∞ i q i

R0 = lim

since

lim

1

n→∞ qn

=0

(b) Annuity in advance

Present Value of Annuity

R0 = r ·

q · (qn − 1) i · qn

with i > 0

235

7.5 Sinking Fund Calculation  n  r·q q −1 · i qn  n  r·q q 1 R0 = · n− n i q q   1 r·q R0 = · 1− n i q R0 =

if q > 1 and i > 0 :   r·q 1 r·q · 1− n = n→∞ i q i

R0 = lim

since

lim

1

n→∞ qn

=0

Example: Mrs. Penny owns a property that she wants to sell. She receives a net annual income of $30,000 for renting out this property. At what price should Mrs. Penny sell the property, assuming an interest rate of 2 % p.a., if she assumes that alternatively the property would belong to her or her family eternally? R0 =

$30, 000 · 1.02 = $1, 530, 000 0.02

The “perpetual” value of the property is $1,530,000 assuming that the interest rate is 2 % p.a. on a permanent, i.e. “perpetual” basis.

7.5 Sinking Fund Calculation The sinking fund calculation deals with the repayment of loans, credits and mortgages. The debt is repaid in instalments within a period agreed upon in advance. The creditor expects interest to be paid.

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7.5.1 Fundamental Terms

Annuity Ak

Total payment per period with k = 1, ..., n. Each annuity consists of an interest rate and a repayment instalment. Ak = Tk + Zk

Repayment Instalment Tk

Amount that reduces the debt at the end of the period by payment, with k = 1, ..., n. Note: The principle amount is only reduced by the repayment instalment.

Interest Amount Zk

Interest for the respective residual debt Ck , with k = 1, ..., n. To determine the interest for annual payments in arrears, the interest rate is multiplied by the residual debt of the previous year. Z = i ·Ck−1

Initial Debt C0

Equivalent to the original loan amount. If n is the total number of all repayment periods, the initial debt C0 is equal to the total of all repayment instalments Tk . n

C0 = T1 + T2 + ... + Tn−1 + Tn = ∑ Tk k=1

237

7.5 Sinking Fund Calculation

Residual Debt Ck

Equivalent to the remaining amount after k periods, with k = 1, ..., n. The residual debt of the current period is equivalent to the difference between the residual debt amount of the previous period and the repayment instalment of the current period. Ck = Ck−1 − Tk The residual debt Ck after k periods is equal to the initial debt C0 minus the sum of the repayment instalments Tk . k

Ck = C0 −(T1 +T2 +...+Tk ) = C0 − ∑ Tk k=1

Period k

Duration for which an annuity Ak is to be paid.

Repayment Period n

Total duration of the loan.

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7.5.2 Annuity Repayment With the annuity repayment, the amount of the annuity is constant during the entire period.

A

constant annuity, A1 = A2 = ... = An = A

The interest portion decreases from period to period, while the repayment portion increases by a corresponding amount. The payment of annuities is equivalent to an annuity in arrears. C0 is the present value of all annuities.

Annuity

A = C0 ·

qn · (q − 1) or A = T1 · qn qn − 1 | {z } annuity factor in arrears

Initial Debt

C0 = A ·

qn − 1 qn · (q − 1) | {z }

present value of annuity factor in arrears

Residual Debt

Ck = C0 · qk − A ·

or C0 = T1 ·

qn − 1 i } | {z

amount of annuity factor in arrears

k −1 qk − 1 or Ck = C0 − T1 · qq−1 q−1

239

7.5 Sinking Fund Calculation

Repayment Instalment

Tk = C0 ·

i · qk−1 qn − 1

T1 = C0 ·

i or T1 = A · q−n (1 + i)n − 1

or Tk = T1 · (1 + i)k−1 or

Interest Amount Zk = A − T1 · qk−1 or Zk = Ck−1 · i or Zk = A · (1 − qk−1 ) +C0 · i · qk−1 Interest Factor

Repayment Period

qn =

A A = T1 A −C0 · i

    A A log ln T T n= or n = log(q) ln(q)

Example 1: Mrs. Penny has taken out a loan of $200,000. She commits herself to pay interest at the end of each year at a rate of 3.5 % p.a. After 6 years she wants the loan to be completely repaid. To keep the annual paid amount constant, Mrs. Penny chooses the annuity repayment. A = $200, 000 ·

1.0356 · 0.035 = $37, 533.64 1.0356 − 1

Z1 = $200, 000 · 0.035 = $7, 000 T1 = $37, 533.64 − $7, 000 = $30, 533.64

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240 k

Residual Amount at the Beginning of the Year

Interest Amount

Repayment Instalment

Annuity

Residual Amount at the End of the Year

1

200,000.00

7,000.00

30,533.64

37,533.64 169,466.36

2

169,466.36

5,931.32

31,602.32

37,533.64 137,864.04

3

137,864.04

4,825.24

32,708.40

37,533.64 105,155.64

4

105,155.64

3,680.45

33,853.19

37,533.64

71,302.45

5

71,302.45

2,495.59

35,038.05

37,533.64

36,264.40

6

36,264.40

1,269.25

36,264.39

37,533.64

0.01

Note: The difference of $0.01 is due to rounding errors.

Example 2: Your bank grant you a loan of $30,000.00 and agrees with you that the loan is to be repaid in the form of an annuity. The initial repayment rate is 20 % and the annual interest rate is 7 %. a) Calculate the exact term of the loan in years, including two decimal places. b) Create an amortization schedule in form of a redemption plan. a) A = T1 · qn ⇔ T1 = T1 = with and

A qn

6, 000 + 2, 100 (1.07)n 6, 000 = 30, 000 · 0.2 (20%) 2, 100 = 30, 000 · 0.07

⇒ A = 6, 000 + 2, 100 = $8, 100

241

7.5 Sinking Fund Calculation

n =? 8, 100 8, 100 = T1 6, 000   8, 100 ln(1.07)n = ln 6, 000   8, 100 n · ln(1.07) = ln 6, 000 (1, 07)n =

n =

ln(8, 100/6, 000) ≈ 4.436 years ln(1.07)

b) k

Account Balance at the Beginning of the Year

Interest Amount

Repayment Instalment

Annuity

Account Balance at the End of the Year

1

30,000

2,100

6,000

8,100

24,000

2

24,000

1,680

6,420

8,100

17,580

3

17,580

1,230.60

6,869.40

8,100

10,710.60

4

10,710.60

749.74

7,350.26

8,100

3,360.34

5

3,360.34

235.22

3,360.34

3,595.56

0

7.5.3 Repayment by Instalments In the case of repayment by instalments, a debt C0 is repaid by repayment instalments T that remain constant each year.

T

constant repayment instalment, T1 = T2 = ... = Tn = T

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The interest amount to be paid decreases from period to period. C0 n

Repayment Instalment

T =

Initial Debt

C0 = n · T

Residual Debt

Ck = C0 − k · T or   k Ck = C0 · 1 − n

Interest Amount Zk = i ·Ck−1 or h i Zk = i · C0 − (k − 1) · T or 

 k−1 Zk = i ·C0 · 1 − n   k−1 C0 Ak = i ·C0 · 1 − + n n | {z } |{z}

Annuity

Zk

Ak =

Example:

Tk

i C0 h · 1 + (n − k + 1) · i n

Mrs. Penny has a loan of $150,000 at an interest rate of 2.25 % p.a. payable in arrears. She agrees to repay the loan in instalments over 5 years.

T =

$150, 000 = $30, 000 5

243

7.5 Sinking Fund Calculation Z1 = $150, 000 · 0.0225 = $3, 375

k

Residual Amount at the Beginning of the Year

Interest Amount

Repayment Instalment

Annuity

Residual Amount at the End of the Year

1

150,000.00

3,375.00

30,000.00

33,375.00 120,000.00

2

120,000.00

2,700.00

30,000.00

32,700.00

90,000.00

3

90,000.00

2,025.00

30,000.00

32,025.00

60,000.00

4

60,000.00

1,350.00

30,000.00

31,350.00

30,000.00

5

30,000.00

675.00

30,000.00

30,675.00

0.00

7.5.4 Repayment with Premium If the repayment amount of a loan exceeds the nominal amount, it is referred to as a premium or agio a. The premium refers to the respective repayment period and is expressed as a fixed percentage α. It is also to be repaid. However, no interest may be paid on the premium, as the creditor only receives interest on the nominal amount of the loan. a

premium or agio

α

premium percentage

The opposite of an agio is a disagio (see chapter 7.5.5).

7.5.4.1 Annuity Repayment with Premium (a) Annuity repayment with non-included premium The premium or agio is paid in addition to the annuity. The annuity

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(= sum of repayment instalment and interest amount) remains constant over the entire period. The premium or agio increases from year to year, as it is calculated as a percentage of the repayment instalment, which also increases. As a result, the annuity including premium (= sum of repayment instalment, interest amount and premium) increases.

Aα

annuity including premium

Example:

Mrs. Penny owes the bank a loan of $100,000, which is to be paid off in 5 years at an interest rate of 1.5 % p.a. through annuity repayment. The bank charges Mrs. Penny with a premium in form of an agio of 5 %.

A = C0 ·

qn · i qn − 1

= $100, 000 ·

T1 = C0 ·

1.0155 · 0.015 = $20, 908.93 1.0155 − 1

qk−1 · i qn − 1

= $100, 000 ·

1.0150 · 0.015 = $19, 408.93 1.0155 − 1

Z1 = Ck−1 · i = $100, 000 · 0.015 = $1, 500 alternatively: Z1 = A − T1 = $20, 908.93 − $19, 408.93 = $1, 500

C1 = C0 − T1 ·

qk − 1 q−1

245

7.5 Sinking Fund Calculation = $100, 000 − $19, 408.93 ·

alternatively: C1 = C0 · qk − A ·

1.0151 − 1 = $80, 591.07 1.015 − 1 qk − 1 q−1

= $100, 000 · 1.0151 − $20, 908.93 · ·

1.0151 − 1 1.015 − 1

= $80, 591.07 a1 = T1 · α = $19, 408.93 · 0.05 = $970.45 Aα = A + a1 = $20, 908.93 + $970.45 = $21, 879.38

k Residual Interest Repayment Annuity Premium Annuity Amount Amount Instalment including at the Premium Beginning of the Year 1 100,000.00 1,500.00

19,408.93 20,908.93

970.45 21,879.38

2

80,591.07 1,208.87

19,700.06 20,908.93

985.00 21,893.93

3

60,891.01

913.37

19,995.56 20,908.93

999.78 21,908.71

4

40,895.45

613.43

20,295.50 20,908.93 1,014.78 21,923.71

5

20,599.95

309.00

20,599.93 20,908.93 1,029.99 21,938.92

Note: Differences are due to rounding errors.

(b) Annuity repayment with included premium The premium or agio is already included in the annuity. This means that the annual amount to be paid remains constant throughout the entire period. One speaks of an annuity with included premium.

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Aα

annuity with premium included

Cα

replacement capital (= redemption value, notional debt)

iα

Replacement interest rate (= notional interest rate), which, when applied to Cα , results in the same amount of interest as the interest rate i charged on the initial capital C0 .

qnα

notional accumulation factor qnα = (1 + iα )n

Annuity

Aα =

i · (1 + iα )n ·C0 or (1 + iα )n − 1

Aα = (1 + iα )n ·Cα ·

(1 + iα ) − 1 (1 + iα )n − 1

i 1+α

Notional Interest Rate

iα =

Redemption Value

Cα = C0 · (1 + α) =

Repayment Instalment with Premium

Tk = Aα − i ·Ck

Repayment Instalment without Premium

Tk =

Aα − i ·Ck 1+α

C0 · i iα

247

7.5 Sinking Fund Calculation Example:

Mrs. Penny owes the bank a loan of $100,000, which is to be paid off in 5 years at an interest rate of 1.5 % p.a. through annuity repayment. The premium or agio of 5 % is to be included in the repayment instalment, so that the total annual payment remains constant during the period.

iα =

Aα =

0.015 = 0.01428571429 1 + 0.05 0.015 · (1 + 0.01428571429)5 · $100, 000 (1 + 0.01428571429)5 − 1

= $21, 908.51

T1 =

$21, 908.51 − 0.015 · $100, 000 = $19, 436.68 1 + 0.05

Z1 = $100, 000 · 0.015 = $1, 500.00

a1 = $21, 908.51 − $19, 436.68 − $1, 500.00 = $971.83

k Residual Interest Repay- Premium RepayAnnuity Amount Amount ment Inment In- including at the Bestalment stalment Premium ginning of without with the Year Premium Premium 1

100,000 1,500.00 19,436.68

971.83 20,408.51 21,908.51

2

80,563.32 1,208.45 19,714.34

985.72 20,700.06 21,908.51

3

60,848.98

912.73 19,995.98

999.80 20,995.78 21,908.51

4

40,853.00

612.80 20,281.63 1,014.08 21,295.71 21,908.51

5

20,571.37

308.57 20,571.37 1,028.57 21,599.94 21,908.51

7 Financial Mathematics

248

7.5.4.2 Repayment of an Instalment Debt with Premium In the case of repayment by instalments, the repayment amount remains constant over the period. The annuity increases by the premium or agio. C0 ·α = Tk ·α n

Premium

a=

Repayment Amount including Premium

Tα = (1 + α) · T k

"

Annuity including Premium

#   1 k−1 α Ak = C0 · + 1− ·i+ or n n n

  C0 C0 k−1 + Ak = C0 · i · 1 − + ·α n n n{z } |{z} | | {z } Z

T

a

Ak = Zk + T k + (1 + α) Ak = Zk + T k + a

Example:

Mrs. Penny owes the bank a loan of $100,000, which is to be paid off in 5 years by means of repayment in instalments at an interest rate of 1.5 % p.a. The bank charges Mrs. Penny a premium in form of an agio of 5 %. a = $20, 000 · 0.05 = $1, 000

249

7.5 Sinking Fund Calculation k

Residual Amount at the Beginning of the Year

Interest Amount

Repayment Premium Instalment

Annuity including Premium

1

100,000.00

1,500.00

20,000.00

1,000.00

22,500.00

2

80,000.00

1,200.00

20,000.00

1,000.00

22,200.00

3

60,000.00

900.00

20,000.00

1,000.00

21,900.00

4

40,000.00

600.00

20,000.00

1,000.00

21,600.00

5

20,000.00

300.00

20,000.00

1,000.00

21,300.00

7.5.5 Repayment with Discount (Disagio) A discount (disagio or debt discount) corresponds to a deduction from the face value, which may be agreed in the case of a loan or the issue of a security.

b

discount or disagio

β

discount rate in percent

The opposite of a disagio is the agio (see chapter 7.5.4).

250

7 Financial Mathematics

Annuity Repayment with Discount The amount paid out is reduced by the amount b if a disagio is agreed on. However, the discount, b, is fully included in the the full amount of the repayment amount. The disagio can be interpreted quasi as interest paid in advance. A disagio reduces the nominal interest rate for the periodic instalment during the entire period of the loan. A disagio can take the form of the amount paid or the issue price. (e.g., when securities are issued) in monetary units (e.g. $) or as a percentage of the loan amount: 5 %. disagio, 95 % amount paid or issue price. For a loan with a disagio, only the loan amount reduced by the disagio is disbursed, not the full loan amount. For example, with a loan amount of $100,000 and a disagio of 5 %, the amount paid is $95,000. However, the repayment sum is equal to the amount of the total loan sum (100 % =$100,000). Accordingly, this liability must be reported in the balance sheet in the full amount of the total loan sum, in this case over $100,000. Legally, the accounting for a disagio differs internationally. Often, as for example in Germany,16 there is an option to capitalize the disagio for companies required to prepare financial statements,

a) to record the disagio immediately, i.e. at the time the loan is granted, in the full amount as an expense in the income statement, or b) to record prepaid expenses17 within the balance sheet and to distribute it by scheduled depreciation over the entire period of the loan.

16 17

In Germany, the accounting of a disagio is regulated in § 250 para. 3 HGB. In Germany pursuant to sec. § 266 para. 2 C. HGB (“prepaid expenses“).

251

7.5 Sinking Fund Calculation 7.5.5.1 Annuity Repayment with Discount when Immediately Booked as Interest Expense

If the disagio is booked as interest expense immediately, the amount of the discount, b, is attributed to that one period only.

Example: A company takes out a loan for $100,000 with a period of five years. A disagio of β = 5 % is agreed with the lending bank. The loan is to be repaid in five years at an interest rate of 1.5 % p.a. on the amount paid by annuity repayment. Since the company needs the $100,000 in full, the lender increases the principal amount to $105,263.16 due to the disagio.18

A = K0 ·

qn · i = qn − 1

= $100, 000 ·

T1 = K0 ·

1.0155 · 0.015 = $20, 908.93 1.0155 − 1

qk−1 · i = qn − 1

= $100, 000 ·

1.0150 · 0.015 = $19, 408.93 1.0155 − 1

Z1 = Kk−1 · i = $100, 000 · 0.015 = $1, 500 alternatively: Z1 = A − T1 =

18

$100, 000 = $105, 263.16 or $105, 263.16 · 0.95 = $100, 000 0.95

7 Financial Mathematics

252

= $20, 908.93 − $19, 408.93 = $1, 500

K1 = K0 − T1 ·

qk − 1 = q−1

= $100, 000 − $19, 408.93 ·

1.0151 − 1 = $80, 591.07 1.015 − 1

alternatively: K1 = K0 · qk − A · =

qk − 1 = q−1

$100, 000 · 1.0151 − $20, 908.93 · ·

1.0151 − 1 = $80, 591.07 1.015 − 1

β = K0 · β = $100, 000 · 0.05 = $5, 000.00

k

Residual Interest RepayAmount at Amount ment the InstalBeginning ment of the Year

Annuity Discount Expenses Including Discount

1 100,000.00 1,500.00 19,408.93 20,908.93 5,263.16 26,172.09 (amount paid)

2

80,591.07 1,208.87 19,700.06 20,908.93

20,908.93

3

60,891.01 913.37 19,995.56 20,908.93

20,908.93

4

40,895.45 613.43 20,295.50 20,908.93

20,908.93

5

20,599.95 309.00 20,599.93 20,908.93

20,908.93

Differences are due to rounding errors.

253

7.5 Sinking Fund Calculation 7.5.5.2 Annuity Repayment with Discount when a Disagio is Included in Prepaid Expenses

Are the legal requirements for such an action met19 the difference between the loan amount and the amount paid (disagio) is to be amortized by annual scheduled depreciation, which can be spread over the entire annual period of the liability.20 For the example above the following is valid: k

Residual Interest RepayAmount at Amount ment the InstalBeginning ment of the Year

Annuity Discount Expenses (linearly Including distribDiscount uted)21

1

100,000.00 1,500.00 19,408.93 20,908.93 1,052.63 21,961.56

2

80,591.07 1,208.87 19,700.06 20,908.93 1,052.63 21,961.56

3

60,891.01 913.37 19,995.56 20,908.93 1,052.63 21,961.56

4

40,895.45 613.43 20,295.50 20,908.93 1,052.63 21,961.56

5

20,599.95 309.00 20,599.93 20,908.93 1,052.63 21,961.56

Differences are due to rounding errors.

7.5.5.3 Instalment Repayment with Discount when Immediately Booked as Interest Expense In the case of instalment repayment, the repayment amount remains constant over the entire period.

19 20 21

In Germany pursuant to § 250 (3) HGB. In Germany pursuant to § 250 (3) sentence 2 HGB. $5, 263.16 / 5 = $1, 052.63

7 Financial Mathematics

254 Example:

A company takes out a loan at $100,000 with a period of five years. A discount of β = 5 % is agreed with the lending bank. The loan is to be repaid in five years at an interest rate of 1.5 % on the amount paid p.a. through instalment repayments. Since the company needs the $100,000 in full, the lender increases the nominal amount to $105,263.16 due to the disagio.22

k

Residual Amount at the Beginning of the Year

Interest Amount

Repayment Discount Expenses Instalment Including Discount

1

100,000.00

1,500.00

20,000.00

5,263.16

26,500.00

2

80,000.00

1,200.00

20,000.00

21,200.00

3

60,000.00

900.00

20,000.00

20,900.00

4

40,000.00

600.00

20,000.00

20,600.00

5

20,000.00

300.00

20,000.00

20,300.00

Differences are due to rounding errors.

7.5.5.4 Instalment Repayment with Discount when a Disagio is Included in Prepaid Expenses Are the legal requirements for such an action met23 the difference between the loan amount and the amount paid (disagio) is to be amortized by annual scheduled depreciation, which can be spread over the entire annual period of the liability. 24 . 22 23 24

$100, 000 = $105, 263.16 or $105, 263.16 · 0.95 = $100, 000 0.95 In Germany pursuant to § 250 (3) HGB, In Germany pursuant to § 250 (3) sentence 2 HGB

255

7.5 Sinking Fund Calculation For the example above the following is valid: k

Residual Amount at the Beginning of the Year

Interest Amount

Repayment Discount Instalment (linearly distributed)

Expenses Including Discount

1

100,000.00

1,500.00

20,000.00

1,052.63

22,552.63

2

80,000.00

1,200.00

20,000.00

1,052.63

22,252.63

3

60,000.00

900.00

20,000.00

1,052.63

21,952.63

4

40,000.00

600.00

20,000.00

1,052.63

21,652.63

5

20,000.00

300.00

20,000.00

1,052.63

21,352.63

7.5.6 Grace Periods If the credit period n p exceeds the repayment period nr , i.e. n p > nr , this is referred to as a grace period. Within this grace period, only interest is payable on the debt. Repayment is suspended, thus the burden on the borrower for this period is reduced. The following applies:

Tk = 0

np

credit period in years

nr

repayment period in years

if

k ≤ n p − nr

7 Financial Mathematics

256 Example: (1) Grace Periods for Annuity Repayment

Camilla takes out a loan of $50,000. Since she is still in vocational training, she would like to postpone repayment for the first three years. After that she can repay the loan within 5 years. The interest rate is 4 % p.a. n p = 8; nr = 5 The repayment period nr is key to calculating the annuity. A = $50, 000 ·

k

1.045 · 0.04 = $11, 231.36 1.045 − 1

Residual Amount at the Beginning of the Year

Interest Amount

Repayment Instalment

Annuity

1

50,000.00

2,000.00

0.00

2,000.00

2

50,000.00

2,000.00

0.00

2,000.00

3

50,000.00

2,000.00

0.00

2,000.00

4

50,000.00

2,000.00

9,231.36

11,231.36

5

40,768.64

1,630.75

9,600.61

11,231.36

6

31,168.03

1,246.72

9,984.64

11,231.36

7

21,183.39

847.34

10,384.02

11,231.36

8

10,799.37

431.97

10,799.37

11,231.36

(2) Grace Periods for Repayment by Instalments Camilla would like to know what the annual burden would be if she were to agree to an instalment repayment instead of an annuity repayment. Tk =

$50, 000 = $10, 000 5

257

7.5 Sinking Fund Calculation

k

Residual Amount at the Beginning of the Year

Interest Amount

Repayment Instalment

Annuity

1

50,000.00

2,000.00

0.00

2,000.00

2

50,000.00

2,000.00

0.00

2,000.00

3

50,000.00

2,000.00

0.00

2,000.00

4

50,000.00

2,000.00

10,000.00

12,000.00

5

40,000.00

1,600.00

10,000.00

11,600.00

6

30,000.00

1,200.00

10,000.00

11,200.00

7

20,000.00

800.00

10,000.00

10,800.00

8

10,000.00

400.00

10,000.00

10,400.00

7.5.7 Rounded Annuities In some cases, rounded amounts are desired as annuity, e.g. for percentage annuity and for the repayment of bonds. The calculation therefore deviates from the strict ideal of constant annuities over the entire period.

7.5.7.1 Percentage Annuity With percentage annuity, the repayment rate iT in the first year is given as a percentage of the debt C0 . The annuity is calcualted together with the interest rate i. The interest saved in the subsequent years is used for repayment. T = iT ·C0

The repayment amount of the 1st year.

This results in crooked periods. Therefore the period is broken down into two components n1 and n2 . A compensation payment is due be-

7 Financial Mathematics

258

cause the period does not occur in full years. If the compensation payment is paid at the end of the period, it is called a final payment. If it is paid at the beginning of the period, it is called an advance payment.

iT

repayment rate of the 1st year

n1

n1 = int(n)

n2

n2 = n − n1

Annuity

A = C0 = (i + iT ) 

Period Final Payment Advance Payment

 i + iT ln iT n= with n ∈ /Z ln(q)   qn1 − 1 ·q An1 +1 = C0 · qn1 − A · i A1 = C0 · q − A ·

qn1 − 1 i · qn1

Example: Camilla takes out a loan of $120,000. The loan is to be repaid in the form of a percentage annuity repayment. The initial repayment rate is 7 %, the annual interest rate 3 % p.a. C0 = $120, 000; i = 0.03 (3 %); iT = 0.07 (7 %) A = $120, 000 · (0.03 + 0.07) = $12, 000  ln n=

0.03 + 0.07 0.07 ln(1.03)

 = 12.06 years

259

7.5 Sinking Fund Calculation (1) Final Payment Camilla makes the compensation payment at the end of the period.  A13 =

$120, 000 · 1.0312 − $12, 000 ·

 1.0312 − 1 · 1.03 0.03

= $810.56

k

Residual Amount at the Beginning of the Year

Interest Amount

Repayment Instalment

Annuity

1

120,000.00

3,600.00

8,400.00

12,000.00

2

111,600.00

3,348.00

8,652.00

12,000.00

3

102,948.00

3,088.44

8,911.56

12,000.00

4

94,036.44

2,821.09

9,178.91

12,000.00

5

84,857.53

2,545.73

9,454.27

12,000.00

6

75,403.26

2,262.10

9,737.90

12,000.00

7

65,665.36

1,969.96

10,030.04

12,000.00

8

55,635.32

1,669.06

10,330.94

12,000.00

9

45,304.38

1,359.13

10,640.87

12,000.00

10

34,663.51

1,039.91

10,960.09

12,000.00

11

23,703.42

711.10

11,288.90

12,000.00

12

12,414.52

372.44

11,627.56

12,000.00

13

786.96

23.61

786.96

810.56

(2) Advance Payment Camilla makes the compensation payment at the beginning of the period.

7 Financial Mathematics

260 A1 = $120, 000 · 1.03 − $12, 000 · = $4, 151.95

k

Residual Amount at the Beginning of the Year

1.0312 − 1 0.03 · 1.0312

Interest Amount

Repayment Instalment

Annuity

1

120,000.00

3,600.00

551.95

4,151.95

2

119,448.05

3,583.44

8,416.56

12,000.00

3

111,031.49

3,330.94

8,669.06

12,000.00

4

102,362.43

3,070.87

8,929.13

12,000.00

5

93,433.30

2,803.00

9,197.00

12,000.00

6

84,236.30

2,527.09

9,472.91

12,000.00

7

74,763.39

2,242.90

9,757.10

12,000.00

8

65,006.29

1,950.19

10,049.81

12,000.00

9

54,956.48

1,648.69

10,351.31

12,000.00

10

44,605.17

1,338.16

10,661.84

12,000.00

11

33,943.33

1,018.30

10,981.70

12,000.00

12

22,961.63

688.85

11,311.15

12,000.00

13

11,650.48

349.51

11,650.48

12,000.00

7.5.7.2 Repayment of Bonds When companies need to fund investments, it is often the case that the required loan amount cannot be raised by a single creditor. The company then issues bonds so that a large number of creditors can participate. The bonds issued are divided into rounded partial amounts w (e.g. $100, $500, $1,000, $5,000 and $10,000).

261

7.5 Sinking Fund Calculation

The calculated repayment instalments Tk are broken down into round partial amounts (final, rounded repayment instalment Tk∗ and repayment arrears Rk ). Only repayment instalments that can be divided by the partial amounts without a remainder are allowed. Partial repayment of an integer unit of measure is not possible.

w

value per bond

pik

number of bonds repaid

Tk∗

final repayment instalment

Rk

repayment arrears

ak

number of bonds to be repaid

i · qn qn − 1

Annuity

A = C0 ·

Interest Amount

Zk = i ·Ck−1

Preliminary Repayment Instalment

Tk = A − Zk

if k = 1

Tk = A − Zk + (1 + i) · Rk−1

if k > 1



Tk w

Number of Bonds to be Repaid

ak = int

Final Repayment Instalment

Tk∗ = w · ak



7 Financial Mathematics

262

Repayment Arrears

Rk = Tk − Tk∗

Residual Debt

Ck = Ck−1 − Tk∗

Examples: (1) Repayment of Bonds with Equal Denomination In bonds with equal denominations, the individual securities have the same nominal value. C0 = $5, 000, 000; i = 0.06 (6 %); n = 5; 5, 000 bonds; w = $1, 000 per bond

A = $5, 000, 000 ·

0.06 · 1.065 = $1, 186, 982.00 1.065 − 1

Z1 = 0.06 · $5, 000, 000 = $300, 000 T1 = $1, 186, 982.00 − $300, 000 = $886, 982.00  a1 = int

$886, 982 $1, 000

 = 886

Tk∗ = 886 · $1, 000 = $886, 000 C1 = $5, 000, 000 − $886, 000 = $4, 114, 000 R1 = $886, 982 − $886, 000 = $982.00

263

7.5 Sinking Fund Calculation

T2 = $1, 186, 982.00 − $246, 840 + (1 + 0.06) · $982.00 = $941, 182.92 | {z } | {z } | {z } Z2

A

k

Ck−1

Zk

(1+i)·R1

Tk

ak

Tk∗

Rl

Ak

1 5,000,000.00 300,000.00 886,982.00

886

886,000.00 982.00 1,186,000.00

2 4,114,000.00 246,840.00 941,183.00

941

941,000.00 182.92 1,187,840.00

3 3,173,000.00 190,380.00 996,796.00

996

996,000.00 795.90 1,186,380.00

4 2,177,000.00 130,620.00 1,057,206.00 1,057 1,057,000.00 205.66 1,187,620.00 5 1,120,000.00 67,200.00 1,120,000.00 1,120 1,120,000.00 0.00 1,187,200.00

(2) Repayment of Bonds with Unequal Denomination In bonds with unequal denominations, the individual securities have different nominal values.

C0 = $5, 000, 000; i = 0.06 (6 %); n = 5

Denomination:

a) 250 bonds at $10,000 each b) 500 bonds at $5,000 each

7 Financial Mathematics

264

Repayment Schedule with Denomination for Partial Bond a): k

Ck−1

Zk

Tk

ak

Tk∗

Rl

Ak

1 2,500,000.00 150,000.00 443,491.00 44 440,000.00 3,491.00 590,000.00 2 2,060,000.00 123,600.00 473,591.46 47 470,000.00 3,591.46 593,600.00 3 1,590,000.00 95,400.00 501,897.95 50 500,000.00 1,897.95 595,400.00 4 1,090,000.00 65,400.00 530,102.83 53 530,000.00 102.83 595,400.00 5 560,000.00

33,600.00 560,000.00 56 560,000.00

0

593,600.00

250 2,500,000

Repayment Schedule with Denomination for Partial Bond b): k

Ck−1

Zk

Tk

ak

Tk∗

Rl

Ak

1 2,500,000.00 150,000.00 443,491.00 88 440,000.00 3,491.00 590,000.00 2 2,060,000.00 123,600.00 473,591.46 94 470,000.00 3,591.46 593,600.00 3 1,590,000.00 95,400.00 501,897.95 100 500,000.00 1,897.95 595,400.00 4 1,090,000.00 65,400.00 530,102.83 106 530,000.00 102.83 595,400.00 5 560,000.00

33,600.00 560,000.00 112 560,000.00 500 2,500,000

0

593,600.00

265

7.5 Sinking Fund Calculation Total Repayment Schedule for Partial Bounds a) and b):

Aa/b = $2, 500, 000 ·

1.065 · 0.06 = $593, 491.00 1.065 − 1

T1a/b = $593, 491 − $150, 000 = $443, 491.00 T2a/b = $593, 491 − $123, 600 + 1.06 · $3, 491 = $473, 591.46 .. . T5a/b = $593, 491 − $33, 600 + 1.06 · $102.83 = $560, 000.00

 a1a = int

$443, 491 $10, 000



$560, 000 $10, 000



 = 44

a1b = int

$443, 491 $5, 000



$560, 000 $5, 000



= 88

.. .  a5a = int

k

Ck−1

 = 56

Zk

a5b = int

Tk∗

Ak

= 112

ak a)

b)

1 500,000.00 300,000.00 886,000.00 1,186,000.00 44

88

2 4,114,000.00 246,840.00 941,000.00 1,187,840.00 47

94

3 3,173,000.00 190,380.00 996,000.00 1,186,380.00 50 100 4 2,177,000.00 130,620.00 1,057,000.00 1,187,620.00 53 106 5 1,120,000.00 67,200.00 1,120,000.00 1,187,200.00 56 112 935,040.00 5,000,000.00 5,935,040.00 250 500

7 Financial Mathematics

266

7.5.8 Repayment During the Year If there are several repayment and/or interest periods within one year, this is known as repayment during the year.

mr

number of repayment periods per year

mz

number of interest periods per year

N

total period of a loan/credit measured in repayment periods

N1

N1 = n1 · mr

N2

N2 = n2 · mr

7.5.8.1 Annuity Repayment During the Year In the case of annuity repayments during the year, the burden of the repayment instalment and interest amount is constant for each repayment period during the year.

The following applies:

A1 = A2 = A3 = ... = An = A

(a) Equal number of interest and repayment periods (mz = mr )

i mz

Relative Interest Rate

j=

Annuity

A = C0 ·

j · (1 + j)N (1 + j)N − 1

267

7.5 Sinking Fund Calculation

N = n · mr

Period of a Loan

Example:

(measured in repayment periods)

Camilla takes out a loan of $12,000 at an interest rate of 5 % p.a. Interest and repayment payments are made quarterly. The loan is expected to be repaid after three years. C0 = $12, 000; i = 0.05 (5 %); n = 3; mz = mr = 4 N = 3 · 4 = 12 quarters (repayment periods) j=

0.05 = 0.0125 4

A = $12, 000 ·

0.0125 · (1 + 0.0125)12 = $1, 083.10 (1 + 0.0125)12 − 1

7 Financial Mathematics

268

Year

Quarter

1

2

3

Residual Amount at the Beginning of the Year

Interest Repayment Amount Instalment

Annuity

1

12,000.00

150.00

933.10

1,083.10

2

11,066.90

138.34

944.76

1,083.10

3

10,122.14

126.53

956.57

1,083.10

4

9,165.57

114.57

968.53

1,083.10

1

8,197.04

102.46

980.64

1,083.10

2

7,216.40

90.21

992.89

1,083.10

3

6,223.51

77.79

1,005.31

1,083.10

4

5,218.20

65.23

1,017.87

1,083.10

1

4,200.33

52.50

1,030.60

1,083.10

2

3,169.73

39.62

1,043.48

1,083.10

3

2,126.25

26.58

1,056.52

1,083.10

4

1,069.73

13.37

1,069.73

1,083.10

(b) More interest than repayment periods (mz > mr )

Relative Interest Rate

 mz  i mr j = 1+ −1 mz

Annuity

A = C0 ·

Period of a Loan

N = n · mr

j · (1 + j)N (1 + j)N − 1 (measured in repayment periods)

7.5 Sinking Fund Calculation Example:

269

Camilla takes out a loan of $12,000 at an interest rate of 5 % p.a. Interest is paid monthly, while the repayment is made quarterly. The loan should be repaid after three years. C0 = $12, 000; i = 0.05 (5 %); n = 3; mr = 4; mz = 12 N = 3 · 4 = 12 quarters (repayment periods)

j=

  12 0.05 4 1+ − 1 = 0.0126 12

A = $12, 000 ·

0.0126 · (1 + 0.0126)12 = $1, 083.78 (1 + 0.0126)12 − 1

7 Financial Mathematics

270 Year

1

2

3

Quarter Residual Interest Repayment Amount Amount Instalment at the Beginning of the Year

Annuity

1

12,000.00

151.20

932.58

1,083.78

2

11,067.42

139.45

944.33

1,083.78

3

10,123.09

127.55

956.23

1,083.78

4

9,166.86

115.50

968.28

1,083.78

1

8,198.58

103.30

980.48

1,083.78

2

7,218.10

90.95

992.83

1,083.78

3

6,225.27

78.44

1,005.34

1,083.78

4

5,219.93

65.77

1,018.01

1,083.78

1

4,201.92

52.94

1,030.84

1,083.78

2

3,171.08

39.96

1,043.82

1,083.78

3

2,127.26

26.80

1,056.98

1,083.78

4

1,070.28

13.49

1,070.28

1,083.77

Note: Differences are due to rounding errors.

(c) More repayment than interest periods (mr > mz ) If there are more repayment periods than interest periods, interest payments no longer accrue at the end of each repayment period, but only if • an interest period has ended. • the end of the period has been reached.

271

7.5 Sinking Fund Calculation Calculated interest at the end of the kth repayment period Zk∗ =

i ·Ck−1 mr

Interest payments at the end of a repayment period

Zk =

       0       

k

∗

Zτ   τ = k−mr +1       N     ∑ Zτ∗ 

∑

τ = n1 ·mr +1

if

k mr

if

k mr

if

k mr

    k   ̸= int and k < N    mr        k = int  mr           k   ̸= int and k = N   mr 



Annuity A = C0 ·

qn1 · (1 + n2 · i) mr + 2i · (mr − 1) ·


qn1 −1 i

  2 · (1 + n2 · i) + N2 + mir · (N2 −1)·N 2

Period components n1 = int(n)

n2 = n − n1

N1 = n1 · mr

N2 = n2 · mr

Example: Nawid, Camilla’s cousin, takes out a loan of $12,000 at an interest rate of 5 % p.a. Repayments are made quarterly while interest is paid annually. The loan is scheduled to be repaid after 2.5 years.

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272

C0 = $12, 000; i = 0.05 (5 %); n1 = 2; n2 = 0.5; mr = 4; mz = 1 N1 = 2 · 4 = 8 quarters (repayment periods) N2 = 0.5 · 4 = 2 quarters (repayment periods) N = 2.5 · 4 = 10 quarters (repayment periods)

A = $12, 000 ·

= $12, 000 ·

Z1∗ =

1.052 · (1 + 0.5 · 0.05)  
1.052 −1 0.05 (2−1)·2 4 + 0.05 2 · (4 − 1) · 0.05 · (1 + 0.5 · 0.05) + 2 + 4 · 2

1.1300625 = $1, 282.33 10.57509375

0.05 · $12, 000 = $150 4

Z2 = 0, since

2 ̸= int 4

  2 and 2 < 10 4

Z4 = $150 + $133.97 + $117.94 + $101.91 = $503.82 |{z } | {z } | {z } | {z } Z1∗

4 since = int 4

Z2∗

Z3∗

Z4∗

  4 4

Z8 = $92.18 + $76.15 + $60.12 + $44.09 = $272.54 | {z } | {z } | {z } | {z } Z5∗

since

8 = int 4

Z6∗

  8 4

Z7∗

Z8∗

273

7.5 Sinking Fund Calculation Z10 = $31.47 + $15.44 = $46.91, | {z } | {z } Z9∗

∗ Z10

10 ̸= int since 4



10 4

 and k = N (10 = 10)

Year Quarter Residual Amount at the Beginning of the Year

Interest

calcul.

1

2

3

Repay- Annuity ment Instalment

payment

1

12,000.00

150.00

0.00

1,282.33 1,282.33

2

10,717.67

133.97

0.00

1,282.33 1,282.33

3

9,435.34

117.94

0.00

1,282.33 1,282.33

4

8,153.01

101.91

503.82

778.51 1,282.33

1

7,374.50

92.18

0.00

1,282.33 1,282.33

2

6,092.17

76.15

0.00

1,282.33 1,282.33

3

4,809.84

60.12

0.00

1,282.33 1,282.33

4

3,527.51

44.09

272.54

1,009.79 1,282.33

1

2,517.72

31.47

0.00

1,282.33 1,282.33

2

1,235.39

15.44

46.91

1,235.39 1,282.30

Note: Differences are due to rounding errors.

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274

7.5.8.2 Repayment by Instalments During the Year In the case of repayment by instalments during the year, the repayment instalments remain constant for each repayment period during the year.

The following is valid:

T1 = T2 = T3 = ... = Tn = T

C0 N

Repayment Instalment

T =

Period of a Loan

N = n · mr

(measured in repayment periods)

(a) Equal number of interest and repayment periods (mr = mz )

Relative Interest Rate

Example:

j=

i mz

Camilla takes out a loan of $12,000 at an interest rate of 5 % p.a. The loan has to be paid off in the form of an instalment repayment. The interest payments and repayments are made quarterly. The loan is scheduled to be repaid after three years. C0 = $12, 000; i = 0.05 (5 %); n = 3; mr = mz = 4 N = 3 · 4 = 12 quarters (repayment periods)

T =

$12, 000 = $1, 000 12

j=

0.05 = 0.0125 12

275

7.5 Sinking Fund Calculation Year

Quarter

1

2

3

Residual Amount at the Beginning of the Year

Interest Amount

Repayment Instalment

Annuity

1

12,000.00

150.00

1,000.00

1,150.00

2

11,000.00

137.50

1,000.00

1,137.50

3

10,000.00

125.00

1,000.00

1,125.00

4

9,000.00

112.50

1,000.00

1,112.50

1

8,000.00

100.00

1,000.00

1,100.00

2

7,000.00

87.50

1,000.00

1,087.50

3

6,000.00

75.00

1,000.00

1,075.00

4

5,000.00

62.50

1,000.00

1,062.50

1

4,000.00

50.00

1,000.00

1,050.00

2

3,000.00

37.50

1,000.00

1,037.50

3

2,000.00

25.00

1,000.00

1,025.00

4

1,000.00

12.50

1,000.00

1,012.50

(b) More interest than repayment periods (mz > mr )

Relative Interest Rate

Example:

j=

  mz i mr 1+ −1 mz

Camilla takes out a loan of $12,000 at an interest rate of 5 % p.a. The loan is to be paid off in the form of an instalment repayment. The interest payments are made monthly while the repayments are made quarterly. The loan is to be repaid after three years. C0 = $12, 000; i = 0.05 (5 %); n = 3; mr = 4; mz = 12

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276

N = 3 · 4 = 12 quarters (repayment periods)

T =

j=

Year

1

2

3

Quarter

$12, 000 = $1, 000 12   12 0.05 4 1+ − 1 = 0.01255 12

Residual Interest Amount Amount at the Beginning of the Year

Repayment Annuity Instalment

1

12,000.00

150.60

1,000.00

1,150.60

2

11,000.00

138.05

1,000.00

1,138.05

3

10,000.00

125.50

1,000.00

1,125.50

4

9,000.00

112.95

1,000.00

1,112.95

1

8,000.00

100.40

1,000.00

1,100.40

2

7,000.00

87.85

1,000.00

1,087.85

3

6,000.00

75.30

1,000.00

1,075.30

4

5,000.00

62.75

1,000.00

1,062.75

1

4,000.00

50.20

1,000.00

1,050.20

2

3,000.00

37.65

1,000.00

1,037.65

3

2,000.00

25.10

1,000.00

1,025.10

4

1,000.00

12.55

1,000.00

1,012.55

277

7.5 Sinking Fund Calculation (c) More repayment than interest periods (mr > mz )

If there are more repayment periods than interest periods, the interest payments no longer accrue at the end of each repayment period. Interest payments only accrue at the end of each repayment period if • an interest period has ended or • the end of the period is reached. Calculated interest at the end of the kth repayment period Zk∗ =

i ·Ck−1 mr

Interest payments at the end of a repayment period

Zk =

       0       

k

∗

Zτ   τ = k−mr +1       N    Zτ∗  ∑ 

∑

τ = n1 ·mr +1

Example:

if

k mr

if

k mr

if

k mr

    k   ̸= int and k < N    mr        k = int  mr           k  ̸= int and k = N    mr 



Camilla takes out a loan of $12,000 at an interest rate of 5 % p.a. The loan is to be paid off in the form of an instalment repayment. Repayments are made quarterly while interest payments are made annually. The loan is scheduled to be repaid after 2.5 years. C0 = $12, 000; i = 0.05 (5 %); n = 2.5; mr = 4; mz = 1

7 Financial Mathematics

278

N = 2.5 · 4 = 10 quarters (repayment periods)

T =

$12, 000 = $1, 200 10

Z1∗ =

0.05 · $12, 000 = $150 4

Z4 = $150 + $135 + $120 + $105 = $510 | {z} | {z} | {z} | {z} Z1∗

Z2∗

4 since = int 4

Z3∗

Z4∗

  4 4

Z8 = $90 + $75 + $60 + $45 = $270 |{z} |{z} |{z} |{z} Z5∗

since

Z6∗

8 = int 4

Z7∗

Z8∗

  8 4

Z10 = $30 + $15 = $45 |{z} |{z} Z9∗

since

∗ Z10

10 ̸= int 4



10 4

 and k = N (10 = 10)

279

7.6 Investment Calculation Year

Quarter Residual Amount at the Beginning of the Year

Interest

calcul.

1

2

3

Repayment Annuity Instalment

payment

1

12,000.00 150.00

0.00

1,200.00 1,200.00

2

10,800.00 135.00

0.00

1,200.00 1,200.00

3

9,600.00 120.00

0.00

1,200.00 1,200.00

4

8,400.00 105.00 510.00

1,200.00 1,710.00

1

7,200.00

90.00

0.00

1,200.00 1,200.00

2

6,000.00

75.00

0.00

1,200.00 1,200.00

3

4,800.00

60.00

0.00

1,200.00 1,200.00

4

3,600.00

45.00 270.00

1,200.00 1,470.00

1

2,400.00

30.00

0.00

1,200.00 1,200.00

2

1,200.00

15.00

45.00

1,200.00 1,245.00

7.6 Investment Calculation In the investment calculation, alternative investment projects are assessed and compared to each other. For this purpose, the information relevant to the investment decision is condensed into an indicator, thus enabling a recommendation to be made for one of the investment alternatives. A distinction is made between static and dynamic methods:

7 Financial Mathematics

280

Static Methods

Dynamic Methods

• • • •

• Annuity method • Net present value method • Return on investment

Cost comparison calculation Profit comparison calculation Amortisation calculation Return on investment

7.6.1 Fundamental Terms

Present Value of Capital C0

The present value of capital corresponds to the value of an investment at the point in time t0 . It is determined by adding the balances of expected incoming and outgoing payments (= series of payments) discounted at the calculatory interest rate i and related to the valuation time t0 .

Amount of Capital Cn

The amount of capital is the profit or loss caused by an investment. It corresponds to the sum of the balances of incoming and outgoing payments (= series of payments) accrued with the calculatory interest rate i to the final point in time of the investment period.

281

7.6 Investment Calculation Final-Value of Assets Vn

The final-value of assets, unlike the amount of capital, splits the calculatory interest rate into the debit interest rate (for the interest on the capital invested in the investment) and the credit interest rate (for the reinvestment of the returns).

Ba

Present value of expenditures/disbursements

Be

Present value of income/proceeds

Aq

Profit annuity

Aa

Expenditure annuity

Ae

Income annuity

zk

Payment balance of the kth period with k = 1, 2, . . . , n

ek

Income/proceeds of the kth period with k = 1, 2, . . . , n

ak

Expenditures/disbursements of the kth period with k = 1, 2, . . . , n

tk

Point in time of the kth period with k = 1, 2, . . . , n

n

Total period

i

Interest rate, rate of discount

r

Internal rate of return (IRR), internal rate of discount

7 Financial Mathematics

282

idebit

Debit interest rate for the return on capital invested

icredit

Credit interest rate for reinvestment of the returns

Payment Series of an Investment

Point in Time

t0

t1

t2

t3

...

tn−1

tn

Income

e0

e1

e2

e3

...

en−1

en

Expenditures

a0

a1

a2

a3

...

an−1

an

Payment Series e0 − a0 e1 − a1 e2 − a2 e3 − a3 ... en−1 − an−1 en − an Adequate Target Rate The adequate target rate displays the return of the investment sum. In this way, the real interest due for a (nominal shown) loan can be accounted for. With the help of the interest rate, the payment series of an investment is transformed into an indicator, on the basis of which the advantageousness can be evaluated (comparatively). The rate of discount: • is the minimum interest rate used to calculate interest on outstanding amounts; • is normally above the market interest rate; • must be set higher, the higher the risk of an investment is;

283

7.6 Investment Calculation

• is, when using borrowed capital (outside capital) ≥ the interest rate of the loan of the borrowed capital by a possible reinvestor.

7.6.2 Fundamentals of Financial Mathematics Compound Interest Calculation

Accumulation factor

(1 + i)n

Discount factor

1 = (1 + n)−n (1 + n)n

Amount of capital

Cn = C0 · (1 + i)n

Net present value (NPV)

C0 = Cn ·

1 = Cn · (1 + i)−n (1 + i)n

Examples: Accumulation factor/discount factor:

End of Year

Accumulation Factor 8 %

Discount Factor 8 %

1

1.0800

0.9259

2

1.1664

0.8573

3

1.2597

0.7938

4

1.3605

0.7350

7 Financial Mathematics

284

Amount of capital: C0 = $2, 000; i = 0.08; n = 5 Cn = $2, 000 · (1 + 0.08)5 = $2, 938.66 Net present value: Cn = $2, 938.66; i = 0.08; n = 5 C0 = $2, 938.66 · (1 + 0.08)−5 = $2, 000

Annuity Calculation

Annuity present value factor Annuity factor Actual cash value of an annuity

(1 + i)n − 1 (1 + i)n · i 1 (1 + i)n · i = annuity present value factor (1 + i)n − 1 R = C0 ·

1 annuity present value factor

R = C0 ·

(1 + i)n · i (1 + i)n − 1

Present value of an annuity

C0 = R ·

(1 + i)n − 1 (1 + i)n · i

Present value of a perpetuity

C0 =

R i

= C0 · annuity factor

285

7.6 Investment Calculation Examples: Annuity factor/annuity present value factor:

End of Year

Annuity Factor 10 %

Annuity Present Value Factor 10 %

1

1.1000

0.9091

2

0.5762

1.7355

3

0.4021

2.4869

4

0.3155

3.1699

Actual cash value of an annuity

C0 = $4, 500; i = 0.1; n = 4

R = $4, 500 · Present value of an annuity

(1 + 0.1)4 · 0.1 − 1 = $1, 419.62 (1 + 0.1)4

R = $1, 419.62; i = 0.1; n = 4

R = $1, 419.62 · Present value of a perpetuity

(1 + 0.1)4 − 1 = $4, 500 (1 + 0.1)4 · 0.1

R = $1, 419.62; i = 0.1

C0 =

$1, 419.75 = $14, 196.20 0.1

286

7 Financial Mathematics

7.6.3 Methods of Static Investment Calculation In the static investment calculation, the reference to time or to temporal (dynamic) changes is either not considered at all or only incompletely. Cost Comparison Method For the cost comparison method, the option with the lowest costs is recommended. The procedure is used in practice for replacement and rationalisation investments, since the revenues are not or only subordinately relevant for the decision. Profit Comparison Method By definition, the profit comparison method considers not only costs but also revenues (profit = revenues − costs). Here, the option with the highest profit is recommended. Amortisation Calculation (Pay-back Method, Pay-off Method or Pay-out Method) The amortisation calculation focuses on the question of when or within which period of time the invested capital of an investment project will be amortised. The project with the shortest amortisation period is recommended. Profitability Calculation The decision criterion of an evaluation of investment alternative is the period profitability within a certain period of time. The profitability puts the profit in relation to the capital employed. The investment with the highest period profitability is recommended.

7.6.4 Methods of Dynamic Investment Calculation The dynamic methods of investment calculation are based on payment series that are related to a specific point in time, i.e. they are based on a specific date. A payment series corresponds to the balances of proceeds and disbursements.

287

7.6 Investment Calculation

7.6.4.1 Net Present Value Method (Net Present Value, Amount of Capital, Final Asset Value) With the net present value method, the decision criterion of an investment is the capital value (NPV or amount of capital). Payment balance zk = ek − ak Net present value C0 = z0 +

z1 z2 zn + +...+ (1 + i) (1 + i)2 (1 + i)n

= z0 + z1 · (1 + i)−1 + z2 · (1 + i)−2 + . . . + zn · (1 + i)−n Amount of capital

Cn = z0 · (1 + i)n + z1 · (1 + i)n−1 + . . . + zn

Final-Value of Assets

Vn = z0 · (1 + idebit )n + z1 · (1 + icredit )n−1 + +z2 · (1 + icredit )n−2 + . . . + zn

According to the net present value method, an investment is considered appropriate if the capital value (i.e. the net present value, the amount of capital or the final asset value) is greater than or equal to zero (C0 ; Cn ; Vn ≥ 0).

Capital value > 0

⇒ The investment is considered advantageous compared to an alternative (financial) investment or an expected minimum return.

Capital value = 0 ⇒ The investment achieves (at least) the required minimum return. Capital value < 0

⇒ The investment is not appropriate.

7 Financial Mathematics

288 Example:

i = 0.08 (8%); n = 4 periods with the time points t0 to t4

Point in Time Proceeds Disbursements Payment Series

t0

t1

t2

t3

t4

$0 $100, 000 $120, 000 $150, 000 $45, 000 $60, 000 $30, 000 $50, 000 $105, 000 $40, 000 −$60, 000 $70, 000 $70, 000 $45, 000

$5, 000

Net Present Value

C0 = −$60, 000 + $70, 000 · (1.08)−1 + $70, 000 · (1.08)−2 + $45, 000 · · (1.08)−3 + $5, 000 · (1.08)−4

C0 = $104, 226.13 Amount of Capital Cn = −$60, 000 · (1.08)4 + $70, 000 · (1.08)3 + $70, 000 · (1.08)2 + $45, 000 · · (1.08) + $5, 000 Cn = $141, 798.50 The amount of capital corresponds to the net present value of capital, which bears interest (at 8 %) over the entire period: Cn = C0 · qn

289

7.6 Investment Calculation

Cn · discount factor = C0

C0 · accumulation factor = Cn

⇒ $141, 798.50 · 1.08−4

⇒ $104, 226.13 · 1.084

= $104, 226.13

= $141, 798.50

7 Financial Mathematics

290 Final Asset Value Debit interest rate = 10 %

Credit interest rate = 8 %

Capital values (net present value, amount of capital or final asset value) express the respective monetary value, which results from the assumed or expected incoming and outgoing payments of a time series (investment), calculated at an internal rate of discount, at the beginning or end of a considered period.

7.6.4.2 Annuity Method The annuity method expresses the (monetary) valuation of an investment in terms of periods. It compares the present value of the income with the present value of the expenditures. n

Present value of income

Be =

Present value of expenditures

Ba =

ek k k = 0 (1 + i)

∑ n

ak k k = 0 (1 + i)

∑

291

7.6 Investment Calculation

Income annuity

Ae = Be ·

qn · (q − 1) qn − 1

Expenditure annuity

Aa = Ba ·

qn · (q − 1) qn − 1

Profit annuity25

Ag = Ae − Aa (1 + i)n · i or Ag = C0 · |{z} (1 + i)n − 1 {z } NPV | annuity factor in arrears

If the profit annuity Ag is greater than or equal to zero, the investment is advantageous: Ag = Ae − Aa ≥ 0

Profit annuity > 0

⇒ The investment is considered advantageous compared to an alternative (financial) investment or expected minimum return.

Profit annuity = 0

⇒ The investment achieves (at least) the required minimum return.

Profit annuity < 0

⇒ The investment is not appropriate.

Example: i = 0.08 (8%); n = 4 periods with the time points t0 to t4

25

Profit in the sense of surplus revenue.

7 Financial Mathematics

292

Point in Time

t0

Proceeds

Payment Series

Be =

t2

t3

$0 $100, 000 $120, 000 $150, 000

Disbursements

Ba = $60, 000 +

t1

$60, 000

$30, 000

$50, 000 $105, 000

−$60, 000

$70, 000

$70, 000

t4 $45, 000 $40, 000

$45, 000

$5, 000

$30, 000 $50, 000 $105, 000 $40, 000 + + + = $243, 398.30 1.08 1.082 1.083 1.084

$100, 000 $120, 000 $150, 000 $45, 000 + + + = $347, 624.43 1.08 1.082 1.083 1.084

Aa = $243, 398.30 ·

0.08 · (1.08)4 = $73, 487.01 (1.08)4 − 1

Ae = $347, 624.43 ·

0.08 · (1.08)4 = $104, 955.05 (1.08)4 − 1

Ag = $104, 995.05 − $73, 487.01 = $31, 468.04 or alternatively with Ag = C0 ·

(1 + i)n · i (1 + i)n − 1

C0 = −$60, 000 + $70, 000 · (1.08)−1 + $70, 000 · (1.08)−2 + $45, 000 · ·(1.08)−3 + $5, 000 · (1.08)−4

C0 = $104, 226.13

293

7.6 Investment Calculation ⇒ Ag = $104, 226.13 ·

(1 + 0.08)4 · 0.08 = $31, 468.04 (1 + 0.08)4 − 1

The profit annuity of $31,468.04 corresponds to the annual surplus (average annual profit rate) of the investment at an interest rate of 8 %. The investment thus appears to be economically reasonable due to the positive profit annuity.

7.6.4.3 Internal Rate of Return Method The internal rate of return r is determined by an approximation method and compared with the rate of discount i.

Discount Factor for the kth Year q−k = (1 + r)−k

Equation for Calculating the Internal Rate of Return 0 = −z0 +

z1 z2 z3 zn + + +...+ (1 + r)1 (1 + r)2 (1 + r)3 (1 + r)n n

⇒ 0 = −z0 +

zk (1 + r)k k=1

∑

Examples: (1) One Period Case

Point in Time

t0

Payment Series

−$100

t1 $110

7 Financial Mathematics

294 ⇒ Internal rate of return: 0 = −100 +

⇒

110 ⇒ r = 0.1 (10 %) (1 + r)1

110 − 100 profit = = 10 % profit capital 100

(2) Payment Series Over Several Periods

Point in Time

t0

Proceeds Disbursements

t1

t2

$0

$82, 100

$73, 000

$50, 000

$55, 000

$43, 000

Setting up the equation to calculate r: 0 = −$50, 000 +

$82, 100 − $55, 000 $73, 000 − $43, 000 + q q2

Multiplication of the equation by

q2 : 1, 000

  $27, 100 $30, 000 q2 + · 0 = −$50, 000 + q q2 1, 000       q2 $27, 100 q2 $30, 000 q2 0 = −$50, 000 · + · + · 1, 000 q 1, 000 q2 1, 000 0 = −50 q2 + 27.1 q + 30

| ÷ (−50)

295

7.6 Investment Calculation Application of the p/q formula: 2

0 = q − 0.542q − 0.6

⇒ q1/2 ⇔ q1/2

s  −0.542 2 (−0.542) =− ± − (−0.6) 2 2 √ = 0.271 ± 0.073441

⇒ q1 = 0.271 + 0.8206 = 1.0916 ⇒ q2 = 0.271 − 0.8206 = −0.5496 q1 corresponds to an internal rate of return of r ≈ 0.0916 (9.16 %). If this internal rate of return is above the (assumed) discount rate, the investment is advantageous. No internal interest rate can be assigned to q2 because of the negative sign. No interest rate could be assigned to a q−value below 1 either.

Chapter 8

Optimisation of Linear Models By means of the Lagrange method or linear optimisation, the relative extremes (minima or maxima) of a linear (target) function can be determined under restrictive linear constraints. If the constraints are given in the form of an equation, the model can be solved with the help of the Lagrange method. However, if the constraints consist of inequalities, the model can be solvable using a linear programming approach (linear programming, linear optimisation).

8.1 Lagrange Method 8.1.1 Introduction The method of Lagrange multipliers1 is a mathematical procedure that determines the relative extremes of a linear mathematical model (= linear target function and linear constraints) if the constraints are given in the form of equations.

8.1.2 Formation of the Lagrange Function Given is a (target) function

f = f (x1 , x2 , . . . , xn )

xi > 0

with

i = 1, ..., n

for which the local extremes are to be determined. 1

Joseph-Louis Lagrange (1736 - 1813) was an Italian mathematician.

© Springer-Verlag GmbH Germany, part of Springer Nature 2023 F. W. Peren, Math for Business and Economics, https://doi.org/10.1007/978-3-662-66975-4_8

297

8 Optimisation of Linear Models

298

The function f is limited by the constraints

φ = φ j (x1 , x2 , . . . , xn )

j = 1, ..., m

The Lagrange function L = L(x1 , x2 , . . . xn ) additively links the (target) function with the restrictions:

L=

f (x1 , . . . , xn ) + λ1 φ1 (x1 , . . . , xn ) + + λ2 φ2 (x1 , . . . , xn ) + .. . + λm φm (x1 , . . . , xn )

λ j : Lagrange multiplier for the j th constraint

λj ∈ R j = 1, . . . , m m m (degree of the numerator > degree of the denominator) Example:

f (x) =

x2 + x x1 + 6

Characteristics • constraints in the domain • discontinuities • poles • asymptotes

Constraints in the domain All x-values, for which the denominator is equal to 0, must be excluded from the domain of the function. D = R\{zeros of the denominator} Example:

f (x) =

(x − 5) (x + 1)2

Set the denominator to 0:

f (−1) =

p

(x + 1)2

=0

|

⇔

x+1

=0

| −1

⇔

x

= −1

−1 − 5 −6 = 2 (−1 + 1)2 0

→ D = R\{−1}

()

336

9 Functions

Discontinuities Discontinuities in broken rational functions usually refer to poles. Example:

f (x) =

x3 x−1

with

D f = R\{1}

⇒ Pole at x = 1 Exceptions are the so-called removable discontinuities. A discontinuity is removable if it is possible to simplify the corresponding function term. This means that the original function also has a zero at this position, which is not defined. Example: f (x) = ⇔

1 − x2 x2 − x − 2

(1 + x) · (1 − x) (1 − x) = (x − 2) · (x + 1) (x − 2)

⇒

zeros at x = −1 and x = 1 discontinuities at x = −1 and x = 2

⇒

D f = R\{−1; 2}

The discontinuity at x = −1 is removable; i.e. the term (x + 1) can be simplified. However, the initial function f (x) has a remaining discontinuity at x = −1, which has to be excluded from the domain. The discontinuity at x =2 is not removable; i.e. the term cannot be simplified. There is a pole here.

9.2 Classification of Functions

337

The type of pole can be determined by the left and right limits (limes) of the function: lim ≈ f (1.9999) = +∞

x→2−

lim ≈ f (2.0001) = −∞

x→2+

There are four types of poles: Example: f (x) =

5 = 5x−1 with D f = R\{0} x

The pole is at x = 0. In such case, the function graph of f (x) at x = 0 could develop as follows:

These four possibilities of a pole exist. In the given case for f (x) = 5x−1 , the graph displays the same behaviour shown in the first graph on the left.

338

9 Functions

9.2.2 Non-rational Functions power functions/root functions/transcendental functions

9.2.2.1 Power Functions f (x) = axk

with

k ∈ R;

a ∈ R;

Df = R

The independent variable x of the function f = f (x) forms the basis (of the exponential expression).

The Form of Power Functions f (x) = axk

339

9.2 Classification of Functions Example: k is positive: f (x) = x2

normal parabola

f (x) = −x2

normal parabola opened downwards

|a|1

stretched parabola

f1 (x) = x3 ;

f2 (x) = x5

f1 (x) = −x3 ;

f2 (x) = −x5

340

9 Functions

k is negative: even power function ⇒ pole (hyperbola) without change of sign 1 ; x2 1 = 4; x

f (x) = x−2 = f (x) = x−4

odd power function ⇒ pole (hyperbola) with change of sign 1 ; x1 1 = 3; x

f (x) = x−1 = f (x) = x−3

341

9.2 Classification of Functions 9.2.2.2 Root Function √l k f (x) = a xk = ax l

with

k ∈ R; l ∈ N with n ≥ 1; a ∈ R with a ̸= 0; D f = R+ 0

The independent variable x is in the radicand. The root function is the inverse function of the corresponding power function.

Examples: f (x) =

√

1

x = x2

f −1 (x) = y2

√ 5 7 g(x) = 8x 7 = 8 x5

g−1 (x)

r  y 7 = 5 8

342

9 Functions

9.2.2.3 Transcendental Functions Functions that are not algebraic are transcendental functions.

9.2.2.3.1 Exponential Functions f (x) = ax

with

a ∈ R+ , D f = R,

a ̸= 1 C f = R+

The independent variable x is in the exponent.

Characteristics: a>1

f strictly monotonically increasing

0 1 f (x) = ax

→

g(x) = an·x

Stretched towards the y-axis

⇒ function f (x) is multiplied by a number n, with n > 1 f (x) = ax

→

g(x) = n · ax

Stretched towards the x-axis

⇒ exponent is multiplied by a number n, with 0 < n < 1 f (x) = ax

→

g(x) = an·x

Compressed towards the y-axis

⇒ function f (x) is multiplied by a number n, with 0 < n < 1 f (x) = ax

→

g(x) = n · ax

345

9.2 Classification of Functions Reflections

Reflection across the x-axis

f (x) = ax

multiplied by −1 equals:

y = 4 · 2x

→

y = −4 · 2x

g(x) = −ax

346

9 Functions

Reflection across the y-axis

1st possibility

2nd possibility

Form the inverse of the base

Multiply the exponent by −1

f1 (x) = ax

g1 (x) = ax

 x 1 f2 (x) = a

g2 (x) = a−x

347

9.2 Classification of Functions Shift Upward shift

Downward shift

⇒ add a constant c to the function, with c > 0 f (x) = ax →

g(x) = ax + c

Left shift

f (x) = ax →

g(x) = ax − c

Right shift

⇒ add a constant c to the exponent, with c > 0 f (x) = ax →

⇒ add a constant c to the function, with c < 0

g(x) = ax+c

⇒ add a constant c to the exponent, with c < 0 f (x) = ax →

g(x) = ax−c

348

9 Functions

Natural Exponential Function f (x) = ex

Euler’s number

→ e = 2.718281828459 Application •

when modeling continuous growth

•

when modeling continuous decay

Requirements •

within the same time intervals

•

with a constant growth factor

9.2.2.3.2 Logarithmic Functions f (x) = loga x

with

a ∈ R+ ,

a ̸= 1;

D f = R+ ;

Cf = R

read as “logarithm of x to the base a”. The logarithmic function is the inverse function of the corresponding exponential function. Characteristics:

a>1 0 0 g(x) = loga (x) − c

Left shift

Right shift

f (x) = loga

f (x) = loga

⇒ add a constant c to the antilogarithm, with c > 0 g(x) = loga (x + c)

⇒ substract a constant c from the anti-logarithm, with c > 0 g(x) = loga (x − c)

353

9.2 Classification of Functions Stretch/Compression

Stretch toward the y-axis f (x) = loga x ⇒ function is multiplied by a factor n, with n > 1 g(x) = n · loga x

Stretch toward the x-axis f (x) = loga x ⇒ anti-logarithm (x) is multiplied by a factor n, with 0 < n < 1 g(x) = loga (n · x)

Compression toward the y-axis f (x) = loga x ⇒ function is multiplied by a factor n, with 0 < n < 1 g(x) = n · loga x

Compression toward the x-axis f (x) = loga x ⇒ anti-logarithm (x) is multiplied by a factor n, with n > 1 g(x) = loga (n · x)

354

9 Functions

9.2.2.4 Trigonometric Functions (Angle Functions/Circular Functions)

α is an arbitrary - counterclockwise - angle in a circle of which the center is located at the origin of a Cartesian coordinate system. α defines the angle formed by the (boundary) points A and P situated on the circle and the origin 0: ◁ AOP. The (end) point P has the coordinates (u/v), thus α can also be written as ◁ (u/v). Angle α =◁ (u, v) =◁ AOP The so-called special case unit circle occurs when the circle has a radius of one unit, x = f (x) = 1 In general, i.e. for any angle, the following applies:

• f (x) = sin α =

v w

Df = R

C f ∈ [−1, 1]

• f (x) = cos α =

u w

Df = R

C f ∈ [−1, 1]

• f (x) = tan α =

v u

D f = R\{x | x =

• f (x) = cot α =

u v

D f = R\{x | x = kπ}

tan α =

sin α cos α

cot α =

cos α sin α

π 2

+ kπ}

C f ∈ (−∞, ∞) = R

C f ∈ (−∞, ∞) = R

355

9.2 Classification of Functions

On the unit circle (r = 1) applies:

sin α = v

(y-coordinate of P)

cos α = u

(x-coordinate of P)

tan α =

sin α cos α

with

α ̸= (2k + 1) · 90◦ ; k ∈ Z

cot α =

cos α 1 = tan α sin α

with

α ̸= k · 180◦ ; k ∈ Z

The following relationships also apply on the unit circle:

cos2 α + sin2 α = 1 sin α =

√

tan α

± 1 + tan2 α

sin α = cos(90◦ − α) cos α =

1 √ ± 1 + tan2 α

cos α = sin(90◦ − α) tan α =

sin α p ± 1 − sin2 α

tan α = cot(90◦ − α) The sign of the root depends on whether it is located in the positive (positive root) or negative (negative root) area of the coordinate system.

356

9 Functions

In a right-angled triangle applies: (of α)

sin α =

a opposite = c hypotenuse

cos α =

b adjacent = c hypotenuse

tan α =

a opposite = b adjacent

cot α =

b adjacent = a opposite

Representation of the Trigonometric Functions:

f (α) = sin α or f (α) = cos α

357

9.2 Classification of Functions f (α) = tan α or f (α) = cot α

Characteristics of Trigonometric Functions (k ∈ Z)

Domain Codomain

sin α

cos α

R

R

[−1.1]

[−1.1] π 2

tan α R\{x | x =

π 2

cot α + kπ} R\{x | x = kπ}

(−∞, ∞) = R

(−∞, ∞) = R 1 2


+k ·π

Zeros

kπ

Poles

-

-

(2k + 1) · ( π2 )

kπ

Extremes

see below

see below

-

-

Inflection points

k·π

Asymptotes

Periods

π 2

+ kπ

+k·π

kπ

k·π

π 2

+k·π

see below

see below

see below

see below

-

-

(2k + 1) · π2

k·π

see below

see below

π

π

2π

2π

358

9 Functions

Extremes  4k + 1 maxima: sin ·π = 1 with k ∈ Z 2   7π 3π π 5π 9π i.e. ..., − , − , , , , ... 2 2 2 2 2 

sin α:

are maxima of sin α 

 4k − 1 · π = −1 with k ∈ Z 2   5π π 3π 7π 9π , , ... i.e. ..., − , − , − , 2 2 2 2 2

minima: sin

are minima of sin α

cos α:

maxima: cos(2k · π) = 1

with

k∈Z

i.e. {. . . , −4π, −2π, 0, 2π, 4π, . . . } are maxima of cos α

minima: cos(2k · π + 1) = −1

with

k∈Z

i.e. {. . . , −3π, −π, π, 3π, 5π, . . . } are minima of cos α

9.2 Classification of Functions

359

Inflection Points sin α:

k·π i.e. {. . . , −2π, −π, 0, π, 2π, . . . } There is a concave/convex inflection point at π which switches left and right to convex/concave, then back again to concave/convex.

cos α:

π +k·π 2   5 π π π 5 i.e. . . . , − π, − π, , + π, π, . . . 2 2 2 2 2 π which 2 switches left and right to convex/concave, then back again to concave/convex. There is a concave/convex inflection point at

tan α:

k·π i.e. {. . . , −2π, −π, 0, π, 2π, . . . } All inflection points are concave/convex.

cot α:

π +k·π 2   5 3 π 3 5 i.e. . . . , − π, − π, , π, π, . . . 2 2 2 2 2 All inflection points are convex/concave.

360

9 Functions

Asymptotes

tan α:

(2k + 1) ·  i.e.

cot α:

π 2

3 1 π 3 5 . . . , − π, − π, , π, π, . . . 2 2 2 2 2



k·π i.e. {. . . , −2π, −π, 0, π, 2π, . . . } All asymptotes at tan α and cot α run vertically, i.e. are vertical asymptotes.

361

9.2 Classification of Functions Trigonometric Values for Common Angles1 α(◦ ) α (rad) 0◦

0

15◦

π 12

18◦

π 10

30◦

π 6

36◦

π 5

45◦

π 4

54◦

3π 10

60◦

π 3

72◦

2π 5

75◦

5π 12

90◦

π 2

sin α 0 √ √ 1 4 ( 6 − 2) √ 1 4 ( 5 − 1)

1 √ √ 1 4 ( 6 + 2) p √ 1 10 + 2 5 4 √ 1 1 2 2 3 p √ √ 1 10 − 2 5 14 (1 + 5) 4 √ √ 1 1 2 2 2 2 p √ √ 1 1 10 − 2 5 4 (1 + 5) 4 √ 1 1 2 3 2 p √ √ 1 1 10 + 2 5 4 ( 5 − 1) 4 √ √ √ √ 1 1 4 ( 6 + 2) 4 ( 6 − 2)

108◦

3π 5

120◦

2π 3

135◦

3π 4

1 p √ 1 10 + 2 5 4 √ 1 2 3 √ 1 2 2

180◦

π

0

270◦

3π 2

360◦

2π

1

cos α

0 √ 1 4 (1 − 5) − 12 √ − 12 2

tan α

cot α

0 √ 2− 3 p √ 1 25 − 10 5 5 √ 1 3 3 p √ 5−2 5

±∞ √ 2+ 3 p √ 5+2 5 √ 3 p √ 1 25 + 10 5 5

1 p √ 1 25 + 10 5 5 √ 3 p √ 5+2 5 √ 2+ 3

√ 5−2 5 √ 1 3 3 p √ 1 25 − 10 5 5 √ 2− 3

1 p

±∞ 0 p p √ √ − 5 + 2 5 − 15 25 − 10 5 √ √ − 3 − 13 3 −1

−1

−1

0

±∞

−1

0

±∞

0

0

1

0

±∞

Cf. Stratosphere Digital (2020): https://formula.amardesh.com/mathematics/trigonometric-functions-of-common-angles/, accessed 4 June 2020.

362

9 Functions

Phase Shifts of Trigonometric Functions  π sin x + = cos x or sin(x + 90◦ ) 2  π cos x + = − sin x or cos(x + 90◦ ) 2  π tan x + = − cot x or tan(x + 90◦ ) 2  π cot x + = − tan x or cot(x + 90◦ ) 2

= cos x = − sin x = − cot x = − tan x

Relationships between Angle Functions

sin α

cos α

tan α cot α

±

p

sin α

cos α

tan α

cot α

sin α

√ ± 1 − cos2 α

tan α √ ± 1 + tan2 α

1 √ ± 1 + cot2 α

cos α

1 √ ± 1 + tan2 α

cot α √ ± 1 + cot2 α

tan α

1 cot α

1 tan α

cot α

1 − sin2 α

±

p

1 − sin2 α

√ ± 1 − cos2 α cos α

±

p

1 − sin2 α sin α

cos α √ ± 1 − cos2 α

sin α

Conversion for any Arbitrary Angle sin α

cos α

tan α

cot α

90◦ ± α

+ cos α

∓ sin α

∓ cot α

∓ tan α

180◦ ± α

∓ sin α

− cos α

± tan α

± cot α

270◦ ± α

− cos α

± sin α

∓ cot α

∓ tan α

360◦ ± α

± sin α

+ cos α

± tan α

− cot α

9.2 Classification of Functions Periodicity of Trigonometric Functions

sin α = sin (α + k · 2π)

     

    cos α = cos (α + k · 2π)        tan α = tan (α + kπ) cot α = cot (α + kπ)

period 2π

period π

    

Symmetries in Trigonometric Functions sin(−x) = − sin x cos(−x) = + cos x tan(−x) = − tan x cot(−x) = − cot x

363

364

9 Functions

Goniometric Transformations Sums and Differences (α ± β )

sin(α ± β ) = sin α cos β ± cos α sin β

cos(α ± β ) = cos α cos β ∓ sin α sin β

tan(α ± β ) =

sin (α ± β ) tan α ± tan β = 1 ∓ tan α tan β cos (α ± β )

cot(α ± β ) =

cot α cot β ∓ 1 cos (α ± β ) = cot β cot α sin (α ± β )

sin α + sin β = 2 sin

α −β α +β cos 2 2

sin α − sin β = 2 cos

α +β α −β sin 2 2

cos α + cos β = 2 cos

α +β α −β cos 2 2

cos α − cos β = 2 sin

α +β α −β sin 2 2

cos α ± sin α =

√

2 sin(45◦ ± α) =

tan α ± tan β =

sin (α ± β ) cos α cos β

cot α ± cot β =

sin (α ± β ) sin α sin β

√

2 cos(45◦ ∓ α)

365

9.2 Classification of Functions  α Double-Angle and Half-Angle Identities 2α; 2 sin 2α = 2 sin α cos α =

2 tan α 1 + tan2 α

cos 2α = cos2 α − sin2 α = 1 − 2 sin2 = 2 cos2 α − 1 =

tan 2α =

2 tan α 2 = 2 1 − tan α cot α − tan α

cot 2α =

cot α − tan α cot2 α − 1 = 2 cot α 2

sin

α =± 2

r

1 − cos α 2

r

1 + cos α 2

α tan = ± 2

r

1 − cos α 1 − cos α sin α = = 1 + cos α sin α 1 + cos α

α cot = ± 2

r

1 + cos α 1 + cos α sin α = = 1 − cos α sin α 1 − cos α

α cos = ± 2

1 − tan2 α 1 + tan2 α

366

9 Functions

Other Multiple-Angle Identities (n · α)

sin 3α = 3 sin α − 4 sin3 α sin 4α = 8 sin α cos3 α − 4 sin α cos α sin 5α = 16 sin α cos4 α − 12 sin α cos2 α + sin α cos 3α = 4 cos3 α − 3 cos α cos 4α = 8 cos4 α − 8 cos2 α + 1 cos 5α = 16 cos5 α − 20 cos3 α + 5 cos α sin n α = n sin α cosn−1 α −

cos n α = cosn α −

n
2 2 sin

n
3 3 sin α

cosn−3 α +

α cosn−2 α +

tan 3α =

3 tan α − tan3 α 1 − 3 tan2 α

tan 4α =

4 tan α − 4 tan3 α 1 − 6 tan2 α + tan4 α

cot 3α =

cot3 α − 3 cot α 3 cot2 α − 1

cot 4α =

cot4 α − 6 cot2 α + 1 4 cot3 α − 4 cot α

n
4 4 sin

n
5 5 sin

α cosn−5 α + . . .

α cosn−4 α + . . .

9.2 Classification of Functions Products (α · β ) sin α sin β =

1 (cos(α − β ) − cos(α + β )) 2

cos α cos β =

1 (cos(α − β ) + cos(α + β )) 2

sin α cos β =

1 (sin(α − β ) + sin(α + β )) 2

tan α tan β =

tan α + tan β tan α − tan β =− cot α + cot β cot α − cot β

cot α cot β =

cot α − cot β cot α + cot β =− tan α + tan β tan α − tan β

tan α cot β =

tan α + cot β tan α − cot β =− cot α + tan β cot α − tan β

367

368

9 Functions

Powers (α n ) sin2 α =

1 (1 − cos 2 α) 2

cos2 α =

1 (1 + cos 2 α) 2

tan2 α =

1 − cos 2 α 1 + cos 2 α

sin3 α =

1 (3 sin α − sin 3 α) 4

cos3 α =

1 (3 cos α + cos 3 α) 4

sin4 α =

1 (cos 4 α − 4 cos 2 α + 3) 8

cos4 α =

1 (cos 4 α + 4 cos 2 α + 3) 8

sin5 α =

1 (10 sin α − 5 sin 3 α + sin 5 α) 16

cos5 α =

1 (10 cos α + 5 cos 3 α + cos 5 α) 16

sin6 α =

1 (10 − 15 cos 2 α + 6 cos 4 α − cos 6 α) 32

cos6 α =

1 (10 + 15 cos 2 α + 6 cos 4 α + cos 6 α) 32

369

9.2 Classification of Functions Arcus Functions

Arcus functions (=cyclometric functions) are the inverse functions of the trigonometric functions (=angle functions/circular functions). However, this only applies to the so-called principal values, i.e. for certain codomains, since the arcus functions are strictly monotonous and thus uniquely reversible only in certain intervals. The following applies for the principal values: π π ≤y≤ 2 2

sin y = x

with

−

cos y = x

with

0≤y≤π

tan y = x

with

−

cot y = x

with

0 0, the parabola is shifted upwards along the y-axis by e units.

For e < 0, the parabola is shifted downwards along the y-axis by e units.

9.3 Characteristics of Real Functions

393

Shift along the x-axis: f (x) = a · (x − d)2 + e For d > 0, the parabola is shifted to the right along the x-axis by d units.

For d < 0, the parabola is shifted to the left along the x-axis by d units.

394

9 Functions

9.3.4 Continuity If a function f is differentiable in x0 , it is continuous there. Discontinuities See also Chapter 9.2. There are three types of discontinuities:

- infinite discontinuities (poles), - removable discontinuities (singularities), - jump discontinuities.

9.3.5 Infinite Discontinuities An infinite discontinuity (pole) of a broken rational function f = f (x) at x0 always exists when the denominator is equal to zero (singularity): lim f (x) = ±∞

x→x0

A pole cannot be removed.

9.3 Characteristics of Real Functions

395

Example: f (x) = ⇒

1 − x2 (1 + x) · (1 − x) (1 − x) = = 2 x −x−2 (x − 2) · (x + 1) (x − 2) D f = R \ {−1; 2}

The singularity at x = 2 cannot be removed; i.e. it cannot be simplified. There is a pole here. lim f (x) ≈ f (1.9999) = +∞

x→2−

lim f (x) ≈ f (2.0001) = −∞

x→2+

396

9 Functions

9.3.6 Removable Discontinuities A removable discontinuity (singularity) of a broken rational function f = f (x) at x0 always exists when the numerator and denominator simultaneously equal zero (singularity). The discontinuities can be removed by assigning the limit lim f (x) to the discontinuities. x→x0

Example: see above f (x) = ⇒

1 − x2 (1 + x) · (1 − x) (1 − x) = = x2 − x − 2 (x − 2) · (x + 1) (x − 2) D f = R \ {−1; 2}

The singularity at x = −1 can be removed; i.e. it can be simplified. The corresponding limit exists:

2 lim f (x) = − ; 3

x→−1−

lim f (x) = −

x→−1+

2 3

⇒

There is a removable discontinuity here.

Remark: In the original function f (x), the discontinuity remains by definition a singularity, which must be excluded from the domain.

9.3 Characteristics of Real Functions

397

9.3.7 Jump Discontinuities At a jump discontinuity x0 of a function f = f (x), the left and right limits are distinct: lim f (x)

x→x0−

̸=

Example:  −1 f (x) =  2 x

lim f (x) (̸= ±∞) | {z }

x→x0+

cf. pole

for x < 1 for x ≥ 1

⇒ jump discontinuity at x = 1

398

9 Functions

9.3.8 Homogeneity A function f with n independent variables x1 , . . . , xn ; f = f (x1 , . . . , xn ) with D f ⊆ Rn is homogeneous of degree r if λ ≥ 0 applies to each real number: f (λ x1 , . . . , xn ) = λ r · f (x1 , . . . , x2 ) · r = degree of homogeneity

Examples: f (x1 , x2 ) = x12 + 4x1 x2 + 5x22 ⇒

f (λ x1 , λ x2 ) = (λ x1 )2 + 4λ x1 λ x2 + 5(λ x2 )2 = λ 2 x12 + λ 2 4x1 x2 + λ 2 5x22 = λ 2 (x12 + 4x1 x2 + 5x22 ) = λ 2 · f (x1 , x2 )

⇒

f is homogeneous of degree 2.

399

9.3 Characteristics of Real Functions

9.3.9 Periodicity A function f with f = f (x) is periodic with a period T if the following applies: f (x) = f (x ± nT )

with

(x ± nT ) ∈ D f

n ∈ Z∗

T >0

The smallest period is also called the (primitive) period of f . Example: The sine function f (x) = sin x is a periodic function with the period T = 2π: sin x = sin (x + k2π)

with

k∈R

9.3.10 Zeros Zero is the intersection of a function f (x) with the x-axis. ⇒ f (x) = 0 Example:

⇒

p/q formula:

f (x) = x2 − x − 6 = 0 s  1 1 2 x1 /x2 = + ± − +6 2 2 √ x1 = 0.5 + 6.25 = 3 √ x2 = 0.5 − 6.25 = −2

400

9 Functions

9.3.11 Local Extremes

!

necessary condition:

f ′ (x) = 0

sufficient condition:

f ′′ (x) > 0 ⇒ minimum f ′′ (x) < 0 ⇒ maximum

Example: f (x) = x3 − 8x2 + 8x − 3 ⇒

⇒

f ′ (x) = 3x2 − 16x + 8 = 0 ¯ + 2.6667 = 0 f ′ (x) = x2 − 5.3x

p/q formula:

5.3¯ x1 , x2 = ± 2

s −

5.3¯ 2

2 − 2.6667

√ x1 = 2.6667 + 4.4444 = 4.7749 √ x2 = 2.6667 − 4.4444 = 0.5585 y1 = f (4.7749) = x3 − 8x2 + 8x − 3 = −38.3320 y2 = f (0.5585) = x3 − 8x2 + 8x − 3 = −0.8532 The function has extremes at P1 (4.7749 | −38.3320) and P2 (0.5585 | −0.8532). ⇒

f ′′ (4.7749) = 6x − 16 = 12.649 > 0

⇒

minimum P1 (4.7749 | −38.3320)

⇒

f ′′ (0.5585) = 6x − 16 = −12.649 < 0

⇒

maximum P1 (0.5585 | −0.8532)

9.3 Characteristics of Real Functions

9.3.12 Monotonicity The monotonicity defines the slope of a function (monotonic = b uniform). Applies to all x1 , x2 ∈ I (I = interval) with x2 > x1 and I ∈ D f :

f (x1 ) > f (x2 ) ⇒ f is strictly monotonically increasing

f (x1 ) ≥ f (x2 ) ⇒ f is monotonically increasing

f (x1 ) < f (x2 ) ⇒ f is strictly monotonically decreasing

f (x1 ) ≤ f (x2 ) ⇒ f is monotonically decreasing

f (x1 ) = f (x2 ) ⇒ f runs parallel to the x-axis ⇒ the slope is always zero ⇒ the monotonicity is always constant

⇒

The monotonic behaviour changes at the extremes (= minimum or maximum).

⇒

The monotonicity has no relation to the continuity; a step function is for example also monotonically increasing/decreasing.

⇒

There is only one slope between two points; not at one point.

401

402

9 Functions

9.3.13 Concavity and Convexity | Inflection Points

Forms of Curvature •

Convexity exists if the second derivative of a function is greater than zero. ( f ′′ (x) > 0)

•

Concavity exists if the second derivative of a function is smaller than zero. ( f ′′ (x) < 0)

9.3 Characteristics of Real Functions

403

At the inflection points (IP), the curvature behaviour of a function changes from concave to convex or from convex to concave. •

concave/convex inflection point: at the point where the slope is negative, the slope is smallest f ′ (x) = minimal f ′′ (x) = 0 f ′′′ (x) > 0

•

convex/concave inflection point: at the point where the slope is positive, the slope is greatest f ′ (x) = maximal f ′′ (x) = 0 f ′′′ (x) < 0

404

9 Functions

9.3.14 Asymptotes An asymptote is a function that approaches another function without intersecting or touching it. There are four different types of asymptotes: horizontal asymptote

oblique asymptote

vertical asymptote

asymptotic curve

405

9.3 Characteristics of Real Functions 9.3.14.1 Horizontal Asymptotes

Determination of the asymptote using limiting behaviour (in Latin: limes) or by comparing the degree of the numerator (n) and the degree of the denominator (m). degree of the numerator =

degree of the numerator
degree of the denominator +1 e.g.: f (x) = Approach:

x4 − 1 x Step 1: determine the degree of the numerator and of the denominator Step 2: polynomial division Step 3: observation of the limit

Step 1:

x4 − 1 x

Step 2:

(x4 + 0x2 − 1) ÷ (x) = x3 −

Step 3:

lim ( 1 ) x→±∞ x

=0

⇒

1 x

g(x) = x3 (equation of asymptotes)

Key rules denominator = 0 and numerator ̸= 0 n m+1

asymptotic curve

410

9 Functions

9.3.15 Tangent Lines to a Curve A tangent line to a curve is a straight line that touches a function f (x) at a point P0 . The slope of the tangent mtan describes the slope of the function f (x) at a point P0 or at a position x0 . → mtan = f ′ (x0 ) Example: given: f (x) = 3x2 + 1

x0 = 1

to be found: y = m · x + b

(equation of tangent line)

1. Determine the derivative of f (x) ⇒ f ′ (x) = 6x 2. Insert the x0 -value into f (x) to obtain the y0 -value ⇒ y0 = 3 · 12 + 1 y0 = 4 3. Insert the x0 -value into f ′ (x) to obtain m ⇒ f ′ (1) = 6 · 1 f ′ (1) = 6 4. Insert m and y into the general form of the equation of a straight line to obtain b ⇒ y = m·x+b 4 = 6·1+b b = −2 5. Equation of the tangent ⇒ y = 6x − 2

9.3 Characteristics of Real Functions

411

9.3.16 Normal Lines to a Curve A normal line to a curve is perpendicular (orthogonal) to the corresponding tangent at the meeting point with the function f (x). Its slope is equal to the negative reciprocal of the corresponding tangent. ⇒ mnorm = −

1 1 =− ′ mtan f (x0 )

Example: given: f (x) = 3x2 + 1

x0 = 1

to be found: y = m · x + b

(equation of normal line)

1. Determine the derivative of f (x) and the slope of the tangent mtan ⇒ f ′ (1) = 6 = mtan 2. Determine the slope of the normal line ⇒ mnorm =

−1 −1 = mtan 6

3. Insert mnorm and P(1|4) into the form of the equation of a straight line to obtain b 1 ⇒ 4 = − ·1+b 6 b=

25 6

4. Equation of the normal line 1 25 ⇒ y = − x+ 6 6 Note: mnorm · mtan = −1 must always apply.

412

9 Functions

9.4 Exercises Example 1: The water level of the Hudson river in NYC from the 6th to the 12th of January 2011 is modelled realistically by the function 1 h(t) = 2.5t 2 · e− 2 t + 3.5. t is given in days, t = 0 corresponds to January 6th , h(t) is measured in metres. 1. Determine the time when the flood as well as the water level reach their maximum. 2. Determine the times when the water level rises or falls the most.

Subtask 1: Form the derivatives: 1

h′ (t) = e− 2 t (−1.25t 2 + 5t) 1

h′′ (t) = e− 2 t (0.625t 2 − 5t + 5) 1

h′′′ (t) = e− 2 t (−0.3125t 2 + 3.75t − 7.5) 1. Necessary condition

h′ (t) = 0

1

t1 = 0

t2 = 4

∨

1

e− 2 t (−1.25t 2 + 5t)

=0

| e− 2 t ̸= 0 for all t ∈ R

−1.25t 2 + 5t

=0

| factorise t

(−1.25t + 5)

=0

| −5

−1.25t

= −5

| ÷(−1.25)

413

9.4 Exercises 2. Sufficient condition

h′ (t) = 0

h′′ (0) = 5 >0 h′′ (4) ≈ −0.677 < 0

∧

h′′ (t) ̸= 0

local minimum local maximum

Determine the y−values h(4) ≈ 8.9 Answer: The water level of the Hudson river reached its peak of about 8.9 metres on January 10th at 12 a.m.

Subtask 2: 1

h′′ (t) = e− 2 t (0.625t 2 − 5t + 5) h′′ (t) = 0

1. Necessary condition 1

t1,2

1

e− 2 t (0.625t 2 − 5t + 5)

=0

| e− 2 t ̸= 0 for all t ∈ R

0.625t 2 − 5t + 5

=0

|: 0.625

=0

| p/q formula

t 2 − 8t + 8 r 8 8 = ± (− )2 − 8 2 2

2. Sufficient condition

h′′ (t) = 0

| t1 ≈ 6.828

∧

t2 ≈ 1.172

h′′′ (t) ̸= 0

1

h′′′ (t) = e− 2 t (−0.3125t 2 + 3.75t − 7.5) h′′′ (6.828) = 0.1163 > 0

changing point from concave to convex

414

9 Functions

The derivative function h′ has a local minimum at t = 6.828. The water level falls most sharply after about 6 days and 20 hours, i.e. on January 12th at around 8 p.m. h′′′ (1.172) ≈ −1.968 < 0

changing point from convex to concave

The derivative function h′ has a local maximum at t = 1.172. The water level rises most significantly after about one day and 4 hours, i.e. on January 12th at around 4 a.m.

Example 2: High ozone concentration can cause irritation of the respiratory tract, as well as coughing and lung diseases in human beings. Its extent is mainly determined by the duration spent in the contaminated air. According to the forecast for the following day, the ozone concentration in a German city between 7 a.m. (t = 0) and 9 p.m. (t = 14) is measured by the function f with the equation f (t) = 0.06 · (0.25t 4 − 10.6t 3 + 101.2t 2 ) + 55 with 0 ≤ t ≤ 14. 1. Determine the time when the highest ozone concentration in the city is predicted. 2. Determine the times when the ozone concentration in the city increases and decreases the most.

Subtask 1: Form the derivatives: f ′ (t) = 0.06 · (t 3 − 31.8t 2 + 202.4t) f ′′ (t) = 0.06 · (3t 2 − 63.6t + 202.4) f ′′′ (t) = 0.06 · (6t − 63.6)

415

9.4 Exercises 1. Necessary condition

0.06 ̸= 0

t1 = 0

∨

∨

f ′ (t) = 0

0.06 · (t 3 − 31.8t 2 + 202.4t)

=0

(t 3 − 31.8t 2 + 202.4t)

=0

(t 3 − 31.8t 2 + 202.4t)

= 0 | factorise t

t(t 2 − 31.8t + 202.4)

=0

(t 2 − 31.8t + 202.4)

= 0 | p/q formula

t2,3 =

(t 2 − 31.8t + 202.4) = 0 p 15.9 ± (−15.9)2 − 202.4

t2 =

23 > 14;

t3 =

8.8

2. Sufficient condition

f ′ (t) = 0

f ′′ (0) = 12.144 >0 f ′′ (8.8) = −7.4976 < 0

t3 ∈ /D

∧

f ′′ (t) ̸= 0

local minimum local maximum

At the time t = 8.8, the maximum is reached. This corresponds to 3:48 p.m. (0.8h = 0.8 · 60mins = 48mins). The ozone concentration in the city reaches its peak at 3:48 p.m.

416

9 Functions

Subtask 2: 1. Necessary condition

0.06 ̸= 0

∨

f ′′ (t) = 0

0.06 · (3t 2 − 63.6t + 202.4)

=0

(3t 2 − 63.6t + 202.4)

=0

(3t 2 − 63.6t + 202.4)

=0

|: 3

=0

| p/q formula

t 2 − 21.2t +

1, 012 15

r t1,2 =

10.6±

(−10.6)2 −

1, 012 15

t1 ≈ 17.3 > 14 ∈ /D t2 ≈ 3.9

2. Sufficient condition

f ′′ (t) = 0

f ′′′ (3.9) = −2.412 ̸= 0

changing point from convex to concave

∨

f ′′′ (t) ̸= 0

At the time t = 3.9, there is an inflection point. The slope at this time: f ′ (3.9) ≈ 21.9 Examine the slope in the neighbourhood of the time interval [0; 14]: f ′ (0) = 0 f ′ (14) = −39.312 The moment of the strongest increment is t = 3.9, i.e. at 10:54 a.m. The moment of the strongest decrement is t = 14, i.e. at 9 p.m.

Chapter 10

Differential Calculus 10.1 Differentiation of Functions with One Independent Variable 10.1.1 General Difference Quotient = average slope of the function f = f (x) between the points P0 and P or between x0 and x1 . = the quotient (the relation) of the differences ∆ y and ∆ x (Fig. 10.1). ∆y y − y0 f (x0 + ∆ x) − f (xo ) = = ∆x x − x0 ∆x

Fig. 10.1: The Difference Quotient of a Function f = f (x) © Springer-Verlag GmbH Germany, part of Springer Nature 2023 F. W. Peren, Math for Business and Economics, https://doi.org/10.1007/978-3-662-66975-4_10

417

418

10 Differential Calculus

Differential Quotient = derivation (= slope) of the function f = f (x) at the position x0 or at/around the point P0 . df (x0 ) = f ′ (x0 ) dx = lim

∆y ∆x

= lim

f (x0 + ∆ x) − f (x0 ) ∆x

∆ x→0

∆ x→0

The progression from the difference quotient to the differential quotient is shown graphically in Fig. 10.2.

Fig. 10.2: Progression from the Difference Quotient to the Differential Quotient

10.1 Functions with One Independent Variable

419

Remark: ∆y means geographically, that the slope triangle gets progres∆ x→0 ∆ x sively smaller, theoretically so small that ∆ x converts to 0. Finally, the slope triangle at the point of interest P0 is so small that it approximately measures the slope of the function f = f (x) in/around P0 . lim

Derivative Function If the function f = f (x) is differentiable in the entire domain, the derivative function exists (1st derivative of f ).

f ′ (x) =

df dx

with

x ∈ Df

Differential

d f = f ′ (x)dx

with

f = f (x); x ∈ D f

d f is called the differential of the function f .

10 Differential Calculus

420

10.1.2 First Derivative of Elementary Functions

f (x)

f ′ (x)

Remarks

c

0

c = constant

xn

n · xn−1

n∈R

c · xn

c · n · xn−1

1 = x−1 x 1 = x−n xn √ x √ n

x

p

g(x)

1 = −x−2 x2 n − n+1 = (−n) · x−n−1 x 1 √ 2 x 1 1 −1 ·xn n −

⇒ chain rule

ln(x)

1 x

ln(g(x))

⇒ chain rule

loga x

1 x · ln (a)

ex

ex

cx

cx ln(c)

eg(x)

⇒ chain rule

sin x

cos x

a ̸= 1

c>0

a, x > 0

10.1 Functions with One Independent Variable

f (x)

f ′ (x)

cos x

− sin x

cot x

−

arcsin x arccos x arctan x arccot x

1 = −(1 + cot2 x) sin2 x 1 √ 1 − x2 1 −√ 1 − x2 1 √ 1 + x2 1 −√ 1 + x2

sinh x

cosh x

cosh x

sinh x

tanh x coth x arsinh x arcosh x artanh x arcoth x

1 = 1 − tanh2 x cosh2 x 1 − = 1 − coth2 x sinh2 x 1 √ 1 + x2 1 √ 2 x −1 1 √ 1 − x2 1 −√ x2 − 1

Remarks

x ̸= kπ, k ∈ R |x| < 1 |x| < 1

x ̸= 0

x>1 |x| < 1 |x| > 1

421

10 Differential Calculus

422

10.1.3 Derivation Rules

Constant Factor Rule

f (x) = c · g(x)

with

c = constant c ∈ R

f ′ (x) = c · g′ (x) Sum Rule

f (x) = u(x) ± v(x) f ′ (x) = u′ (x) ± v′ (x)

Product Rule

f (x) = u(x) · v(x) f ′ (x) = u′ (x) · v(x) + u(x) · v′ (x)

In general: f (x) = f1 (x) · f2 (x) · . . . · fn (x) f ′ (x) = f1′ (x) · f2 (x) · . . . · fn (x) + f1 (x) · f2′ (x) · . . . · · fn (x) + f1 (x) · f2 (x) · . . . · fn′ (x) Quotient Rule

f (x) =

u(x) v(x)

f ′ (x) =

u′ (x)v(x) − u(x)v′ (x) (v(x))2

with

v(x) ̸= 0

423

10.1 Functions with One Independent Variable

Chain Rule

f (x) = u · v = u(v(x))

with x ∈ Dv Cv ⊂ Du

f ′ (x) = u′ (v(x)) · v′ (x) = outer times inner derivative

In general: f (g1 (g2 (g3 . . . gn (x)))) d f dg1 dg2 dgn df = · · ·...· dx dg1 dg2 dg3 dx

Examples: Constant Factor Rule:

f (x) = 5x20 ⇒ f ′ (x) = 5 · 20x20−1 ⇒ f ′ (x) = 100x19

Sum Rule:

Product Rule:

2 f (x) = 2ex + 4 ln x − √ x 4 1 ⇒ f ′ (x) = 2ex + + √ x x3 f (x) = x7 · ln x ⇒ f ′ (x) = 7x6 · ln x + x7 ·

Quotient Rule:

1 x

ex x4 + 1 ex · (x4 + 1) − ex · (4x3 ) ⇒ f ′ (x) = (x4 + 1)2 f (x) =

10 Differential Calculus

424 Chain Rule:

  1 0.5 f (x) = ln x  −0.5 1 1 ⇒ f (x) = 0.5 · ln · 1 · (−1) · x2 x x ′

10.1.4 Higher Derivations

gen. (recursive) definition

f (n+1) (x) =



f (n) (x)

′

=

d f (n) (x) dx

with n ∈ Z∗
′

d 2 f (x) dx2

2nd derivative

f ′′ (x) = f ′ (x)

3rd derivative


′ d 3 f (x) f ′′′ (x) = f ′′ (x) = dx3

4th derivative

f (4) (x) = f ′′′ (x)

nth derivative

f (n) (x) =



=


′

=

f (n−1) (x)

Examples: (1)

(2)

1 4 1 3 x − x + 2x + 1 4 6 1 f ′ (x) = x3 − x2 + 2 2

f (x) =

′′

2

f (x) = 3x − 1x f ′′′ (x) = 6x − 1

f (x) = lnx f ′ (x) =

1

x

1 x2 1·2 f ′′′ (x) = 3 x ′′

f (x) = −

′

d 4 f (x) dx4 =

d n f (x) dxn

425

10.1 Functions with One Independent Variable

.. .

1·2·3 x4 1·2·3·4 f (5) (x) = x5 .. .

f (n) (x) = 0

f (n) (x) = (−1)n−1 ·

f (4) (x) = 6

f (4) (x) = −

f (5) (x) = 0

(n − 1)! xn

10.1.5 Differentiation of Functions with Parameters f = f (x) indicates that x is the (only) independent variable of the function f . If other placeholders appear in the function term, these are parameters, not variables. Accordingly, f cannot be differentiated with respect to them. Example: f (x) = 2x2 z − z3 + z2 lnx f ′ (x) = 4xz − z2

1 x

f ′ (z) not possible, as z is a parameter.

10.1.6 Curve Sketching An analysis of the function f = f (x) generally includes the complete examination of the following criteria:

Domain D f including discontinuities; check for possible removable discontinuities.

10 Differential Calculus

426 Codomain C f

is usually recommended as the last point of sketching.

Symmetry f (x) = f (−x)

f is axially symmetric to the y-axis

f (x) = − f (−x)

f is point symmetric to the point of origin

Axial Symmetry If the function f , with f = f (x), is axially symmetric to the axis of symmetry x = a, the following applies to all x ∈ D f :

f (a + x) = f (a − x) with a = constant (= axis of symmetry)

If a = 0, a is equivalent to the y-axis. Then the following is valid: f (x) = f (−x) A polynomial function f , whose graph is axially symmetric to the y-axis, has the form f (x) = qn xn + qn−2 xn−2 + ... + q2 x2 + q0 with q = constant and n = even.

427

10.1 Functions with One Independent Variable Point Symmetry

If the function f , with f = f (x), is point symmetric to an arbitrary point P = P(a, b), the following applies for all x ∈ D f :

f (a + x) − f (a) = − f (a − x) + f (a) with a = const. = x-coordinate of P with b = const. = y-coordinate of P

Zeros = Intercepts with the x-Axis !

f (x) = 0

Intercepts with the y-Axis !

x=0

(periodicity)

f (x) = f (x ± n · T )

T = period of f ;

n ∈ Z∗ ;

T >0

Continuity/Differentiability If the function f , with f = f (x), can be differentiated in D f (or in intervals of D f ), it is also continuous here. Incl. discontinuities (poles, gaps, jumps). Pole at x = xP Removable discontinuity at x = xDis

lim f (x); lim f (x)

x→xP−

x→xP+

These discontinuities are removable by definition; with D f = R\{xDis }

Jump at x = xJ

lim f (x) ̸= lim f (x)

x→xJ−

x→xJ+

Clearly allocate limits of intervals

10 Differential Calculus

428 Extrema

f has a relative/local maximum at position x0 , if the following is valid: f ′ (x0 ) = 0

necessary condition

f ′′ (x0 ) < 0

sufficient condition

f has a relative/local minimum at position x0 , if the following is valid: f ′ (x0 ) = 0

necessary condition

f ′′ (x0 ) > 0

sufficient condition

In the extrema the function changes its monotonic behaviour.

Inflection Points f has an inflection point at position x0 , if the following is valid: f ′′ (x0 ) = 0

necessary condition

f ′′′ (x0 ) ̸= 0

sufficient condition

In the inflection points the function changes its curvature behaviour: f ′′′ (x0 ) < 0

convex/concave inflection point

f ′′′ (x0 ) > 0

concave/convex inflection point

Furthermore, if f ′ (x0 ) = 0, then this is a special kind of inflection point, namely a saddle point. The saddle point is an inflection point with a slope of zero and, therefore, it has a horizontal tangent.

429

10.1 Functions with One Independent Variable Monotonicity f ′ (x0 ) ≥ 0

f (in the interval) monotonically increasing

f ′ (x0 ) > 0

f (in the interval) strictly monotonically increasing

f ′ (x0 ) ≤ 0

f (in the interval) monotonically decreasing

f ′ (x0 ) < 0

f (in the interval) strictly monotonically decreasing

Curvature f ′′ (x0 ) > 0

f (in the interval) convex (curvature left)

f ′′ (x0 ) < 0

f (in the interval) concave (curvature right)

p (1 + ( f ′ (x))2 )3 p= f ′′ (x)

Curvature radius:

with

f ′′ (x) ̸= 0

Centre C(xC ; yC ) of the osculating circle with the coordinates: xC = x −

f ′ (x) · (1 + ( f ′ (x))2 ) f ′′ (x)

yC = f (xC ) = f (x) +

(1 + ( f ′ (x))2 ) f ′′ (x)

Behaviour at the Boundaries lim f (x)

left boundary of f

lim f (x)

right boundary of f

x→−∞ x→+∞

10 Differential Calculus

430

The respective optimum optimorum (smallest minimum/ largest maximum) can lie in a relative/ local extremum as well as at one or both boundaries.

Asymptotes An asymptote is a function to which another function increasingly converges. Asymptotes of broken rational functions: n = degree of numerator

m = degree of denominator

n m+1

curved asymptote

Numerator = 0

perpendicular x-axis (at discontinuities/ poles) even multiplicity of the zero ⇒ with change of sign odd multiplicity of the zero ⇒ without change of sign

Graph of the Function The graphic plane f = f (x) is spanned by a two-dimensional Cartesian coordinate system.

431

10.1 Functions with One Independent Variable Example of Curve Sketching: f (x) =

5x − 4 (2 − 3x)2 n o 2 3

Domain:

D f = R\

Codomain:

Cf = ] − ∞ ;

25 24

]

Symmetry: (a) Axial symmetry to the y-axis

f (x) = f (−x) 5x − 4 5(−x) − 4 ̸= (2 − 3x)2 (2 − 3(−x))2

(b) Point symmetry to the origin

f (x) = − f (−x)   5x − 4 5(−x) − 4 ̸= − (2 − 3x)2 (2 − 3(−x))2

⇒ f is neither axially nor point symmetric (also seen in graph of function).

Zeros:

!

f (x) = 0 !

5x − 4 = 0 x=

4 5

⇒ f has a zero at point Z1 Intercepts with the y-Axis:

!

x=0

4 5

|0




10 Differential Calculus

432 5·0−4 −4 = = −1 (2 − 3 · 0)2 4 Periodicity:

f (x) ̸= f (x ± n · T )

with n ∈ Z∗ T > 0

⇒ f is not periodic Continuity:

f is continuous in D f = R\

n o

⇒ f is discontinuous in x =

2 3

as differentiable.

2 3

⇒ pole lim f (x) ≈ f (0.6667) = −∞ x→ 23

+

lim f (x) ≈ f (0.6665) = −∞ x→ 23

Extrema:

−

f (x) =

5x − 4 u(x) = (2 − 3x)2 v(x)

f ′ (x) =

u′ (x)v(x) − u(x)v′ (x) (v(x))2

f ′ (x) =

5 · (2 − 3x)2 − ((5x − 4) · 2(2 − 3x) · (−3)) (2 − 3x)4

=

5 · (2 − 3x) − ((5x − 4) · 2 · (−3)) (2 − 3x)3

=

15x − 14 10 − 15x + 30x − 24 = (2 − 3x)3 (2 − 3x)3

⇒

15x − 14 ! 14 ! = 0 ⇒ 15x − 14 = 0 ⇒ x = 3 (2 − 3x) 15

433

10.1 Functions with One Independent Variable

f ′′ (x) =

15 · (2 − 3x)3 − ((15x − 14) · 3(2 − 3x)2 · (−3)) (2 − 3x)6

=

15 · (2 − 3x) − ((15x − 14) · 3 · (−3)) (2 − 3x)4

=

30 − 45x − (−135x + 126) (2 − 3x)4

=

90x − 96 (2 − 3x)4

1875 14 < 0 ⇒ maximum at x = 64 15

14 14 25 f 15 = 0.452 ⇒ maximum at Emax 15 | 24 f ′′

14 15




Inflection Points: f ′′ (x) =

=−

16 90x − 96 ! ! = 0 ⇒ 90x − 96 = 0 ⇒ x = (2 − 3x)4 15

f ′′′ (x) = f ′′′

16 15




810x − 972 (2 − 3x)5 =

−108 ≈ 43.403 > 0 (− 65 )5

⇒ concave/convex inflection point at IP1

Monotonicity:

] −∞ ;

2 3

] 23 ;

[

14 15

[

] 14 15 ; +∞ [

16 15

|

25 27

f decreases strictly monotonically f increases strictly monotonically f decreases strictly monotonically




10 Differential Calculus

434

Curvature:

Behaviour at the Boundaries:

] − ∞ ; 23 [ ] 23 ; 16 15 [

concave concave

] 16 15 ; +∞ [

convex

lim f (x) ≈ f (1000) = 0+

x→+∞

lim f (x) ≈ f (−1000) = 0−

x→−∞

Asymptotes:

f (x) =

n=1 5x − 4 ⇒ ⇒n 0 ⇒ extremum at point P0 (x0 , y0 , z0 ) with z0 = f (x0 , y0 ) ′′ (x , y ) < 0 if fxx 0 0

and

′′ (x , y ) < 0 ⇒ maximum at point fyy 0 0 P0 (x0 , y0 , z0 )

′′ (x , y ) > 0 if fxx 0 0

and

′′ (x , y ) > 0 ⇒ minimum at point fyy 0 0 P0 (x0 , y0 , z0 )

det H f (x0 , y0 ) < 0 ⇒ saddle point at point P0 (x0 , y0 , z0 ) with z = f (x0 , y0 ) ′′ (x , y ) > 0 and f ′′ (x , y ) < 0 ⇒ saddle point, convex in with fxx 0 0 yy 0 0 x direction and concave in y direction ′′ (x , y ) < 0 and f ′′ (x , y ) > 0 ⇒ saddle point, concave in or fxx 0 0 yy 0 0 x direction and convex in y direction

•

det H f (x0 , y0 ) = 0 ⇒ Indifference, i.e. a decision whether a relative extremum or a saddle point is present at P0 (x0 , y0 , z0 ) is not possible. In this case, either the curvatures in x and y directions are to be measured separately or the function value at P0 with z0 = f (x0 , y0 ), must be compared to neighbouring values.

In other notation, the procedure can be described as follows: •

′′ (x , y ) · f ′′ (x , y ) > ( f ′′ (x , y ))2 fxx 0 0 yy 0 0 xy 0 0

⇒ extremum at point P0 (x0 , y0 , z0 ) with z = f (x0 , y0 ) ′′ (x , y ) < 0 and f ′′ (x , y ) < 0 ⇒ maximum at point if fxx 0 0 yy 0 0 P0 (x0 , y0 , z0 )

10.2 Functions with More Than One Independent Variable

443

′′ (x , y ) > 0 and f ′′ (x , y ) > 0 ⇒ minimum at point if fxx 0 0 yy 0 0 P0 (x0 , y0 , z0 )

•

′′ (x , y ) · f ′′ (x , y ) < ( f ′′ (x , y ))2 fxx 0 0 yy 0 0 xy 0 0

⇒ saddle point at point P0 (x0 , y0 , z0 ) with z0 = f (x0 , y0 ) ′′ (x , y ) > 0 and f ′′ (x , y ) < 0 ⇒ saddle point, convex in with fxx 0 0 0 yy 0 x direction and concave in y direction ′′ (x, y) < 0 and f ′′ (x, y) > 0 ⇒ saddle point, concave in or fxx yy x direction and convex in y direction

•

′′ (x , y ) · f ′′ (x , y ) = ( f ′′ (x , y ))2 fxx 0 0 yy 0 0 xy 0 0

⇒ Indifference, i.e. a decision whether a relative extremum or a saddle point is present at P0 (x0 , y0 , z0 ) is not possible. In this case, either the curvatures in x and y directions are to be measured separately or the function value at P0 with z0 = f (x0 , y0 ), must be compared to neighbouring values. Example 1:

f (x; y) = x3 + 3x2 y − 3xy2 − 21x + y3 − 3y

1st step:

f ′ (x) = 3x2 + 6xy − 3y2 − 21 fy′ = 3x2 − 6xy + 3y2 − 3 ′′ = 6x + 6y fxx ′′ = −6x + 6y fyy ′′ = 6x − 6y fxy !

equation I fx′ (x, y) = 3x2 + 6xy − 3y2 − 21 = 0 !

equation II fy′ (x, y) = 3x2 − 6xy + 3y2 − 3 = 0 Addition method, since all y′ s cancel out.

10 Differential Calculus

444 I II

3x2 + 6xy − 3y2 − 21 3x2 − 6xy + 3y2 − 3 + 6x2 − 24 = 0 6x2 = 24 x2 = 4 ⇒ x1 = 2 and x2 = −2

Insert the values for x1 and x2 in one of the first order derivatives ( f ′ x or f ′ y) to determine the corresponding y-values. (1) for x0 = 2 : (insert in I) 0 = 3 · 22 + 6 · 2y − 3y2 − 21 0 = 12 + 12y − 3y2 − 21 0 = −3y2 + 12y − 9 | : (−3) 0 = y2 − 4y + 3

√ ⇒ 2± 4−3 2±1 y11 = 3 and y12 = 1 ⇒ P1 (2 | 3); P2 (2 | 1)

(2) for x0 = −2 : (insert in I) 0 = 3 · (−2)2 + 6 · (−2)y − 3y2 − 21 0 = 12 − 12y − 3y2 − 21 0 = −3y2 − 12y − 9| : (−3) 0 = y2 + 4y + 3

√ ⇒ −2 ± 4 − 3 −2 ± 1 y21 = −3 and y22 = −1 ⇒ P3 (−2 | − 1); P4 (−2 | − 3)

10.2 Functions with More Than One Independent Variable 2nd step: P1 (2 | 3) ⇒ (6 · 2 + 6 · 3) · (−6 · 2 + 6 · 3) ⪌ (6 · 2 − 6 · 3)2 | {z } | {z } | {z } ′′ (2 | 3) fxx

30

′′ (2 | 3) fyy

·

6

′′ (2 | 3) fxy

>

36

180 > 36

⇒ extremum at P1

⇒ 30 > 0 ; 6 > 0

⇒ minimum at P1

P2 (2 | 1) ⇒ (6 · 2 + 6 · 1) · (−6 · 2 + 6 · 1) < (6 · 2 − 6 · 1)2 −108 < 36

⇒ saddle point at P2

P3 (−2 | − 1) ⇒ [6 · (−2) + 6 · (−1)] · [(−6) · (−2) + 6 · (−1)] < [6 · (−2) − 6 · (−1)]2 −108 < 36

⇒ saddle point at P3

P4 (−2| − 3) ⇒ [6 · (−2) + 6 · (−3)] · [(−6) · (−2) + 6 · (−3)] ⪌ [6 · (−2) − 6 · (−3)]2 108 > 36

⇒ extremum at P4

⇒ −30 < 0 ; −6 < 0

⇒ maximum at P4

445

10 Differential Calculus

446 Example 2:

A manufacturer of bicycles produces two different types A and B of a bicycle. The price of a type A bicycle is $1, 200 per unit and the price of a type B bicycle is $700 per unit. The costs of producing x units of type A and y units of type B are descripted by following cost function: C(x, y) = 150x2 − 100xy + 60y2 − 400x − 500y − 10, 000

a) Determine the production level that maximizes the bicycle manufacturer’s profit. b) What is the maximum profit in $? a) 1. set up profit function P(x, y) = R(x, y) − C(x, y) determine revenue function R(x, y) = 1, 200x + 700y set up profit function P(x, y) = 1, 200x + 700y − (150x2 − 100xy + 60y2 − 400x − 500y − − 10, 000) = = 1, 200x + 700y − 150x2 + 100xy − 60y2 + 400x + 500y + + 10, 000 = = 1, 600x + 1, 200y − 150x2 + 100xy − 60y2 + 10, 000

10.2 Functions with More Than One Independent Variable 2. necessary condition P(x, y) = 1, 600x + 1, 200y − 150x2 + 100xy − 60y2 + 10, 000 P′ x(x, y) = 1, 600 − 300x + 100y = 0 P′ y(x, y) = 1, 200 + 100x − 120y = 0 | · 3 I

1, 600 − 300x + 100y = 0

II

3, 600 + 300x − 360y = 0

I + II 5, 200

(addition method)

− 260y = 0

solve for y 5, 200 − 260y = 0 | + 260y 5, 200 = 260y | : 260 20 = y insert result in I I 1, 600 − 300x + 100 · 20 = 1, 600 − 300x + 2, 000 = 3, 600 − 300x = 3, 600 = 12 =

0 0 0 | + 300x 300x | : 300 x

→ potential extremum at P(12|20)

447

10 Differential Calculus

448 3. sufficient condition P′′ xx = −300 P′′ yy = −120 P′′ xy = 100 G′′ yx = 100

⇒ identical cross derivatives

calculate determinant P′′ xx · P′′ yy − P′′ xy · P′′ yx −300 · (−120) − 100 · 100 > 0 P′′ xx < 0

P′′ yy < 0

→ P(12|20) maximum

The profit is maximal at a production mix of 12 units of type A and 20 units of type B.

b) Calculation of the maximum possible profit Insert point (12|20) in P(x, y) P′ y(x, y) = 1, 600 · 12 + 1, 200 · 20 − 150 · 122 + 100 · 12 · 20 − 60 · 202 + + 10, 000 = = 31, 600 The maximum profit possible is $31, 600.

10.2 Functions with More Than One Independent Variable

449

10.2.3.2 Relative Extrema with m Constraints of the Function f = f (x1 , . . . , xn ) with m < n ⇒

multiplication method according to Lagrange4

previously:

f = f (x1 , . . . , xn )

now:

(target) function + constraints = model

The system of equations (model) to be solved consists of a so-called target function and of one or multiple constraints that limit the solution set of the (target) functions. target function:

f = f (x1 , . . . , xn )

constraints:

g1 = g1 (x1 , . . . , xn ) g2 = g2 (x2 , . . . , xn ) .. . gm = gm (xm , . . . , xn )

precondition:

m 0, ax + b > 0

11 Integral Calculus

462

1 x

ln x + c

x>0

1 ax + b

1 ln(ax + b) + c a

ax + b > 0, a ̸= 0

ex

ex + c

x∈R

eax+b

1 ax+b e +c a

a ̸= 0

sin x

− cos x + c

x∈R

cos x

sin x + c

x∈R

Examples: 1 8 x +c 8

(1) x7 dx = R

R

(2) dx = (3)

R√

R

1dx = x + c

ydy =

R

(4) (2x)4 dx = R

(5)

R dx

√ 5

x2

=

1

y 2 dy =

1 (2x)5 · +c 2 5

R −2 5 3 x 5 dx = x 5 + c

(6) (3z − 2)2 dz = R

2 3 y2 +c 3

3

1 (3z − 2)3 1 · + c = (3z − 2)3 + c 3 3 9

463

11.2 The Indefinite Integral (7)

R√

1 2 1p 3 · · (2x − 1) 2 + c = (2x − 1)3 + c 2 3 3

1

R

2x − 1dx = (2x − 1) 2 dx =

(8) e0.5t dt = 2 · e0.5t + c R

11.2.2 Elementary Calculation Rules for the Indefinite Integral For integrating a function f multiplied by a constant factor and the integration of a sum of two functions f (x) ± g(x), the following rules apply: Let f , g be continuous functions. Then the following is valid: R

R

(1) a · f (x)dx = a · f (x)dx R

(2) ( f (x) ± g(x))dx =

R

R

f (x)dx ± g(x)dx

Examples: (1) (2)

R 1 6x2 dx = 6 x2 dx = 6 · x3 + c = 2x3 + c 3 R 12 (8x3 − 4x + 2 + √ )dx 4x + 9

R

= (8x3 − 4x + 2)dx + √ R

R

12 dx 4x + 9 1

= (8x3 − 4x + 2)dx + 12 (4x + 9)− 2 dx R

R

1 1 = 8 x4 − 4 x2 + 2x + 12 · 4 2 = 2x4 − 2x2 + 2x + 12 ·

1

1 (4x + 9) 2 · 1 4 2

!

1 1 · 2 · (4x + 9) 2 + c 4

√ = 2x4 − 2x2 + 2x + 6 4x + 9 + c

+c

464

11 Integral Calculus

11.3 The Definite Integral 11.3.1 Introduction The other task of integral calculus is to determine the area F of the surface piece, which is bounded by the function graph, the x-axis and the two perpendiculars x = a and x = b. First, the surface area, i.e. the surface measure F, of the gray marked area in the figure below shall be determined. Since not all boundary lines of the grey area are straight, elementary geometric methods fail. Example:

11.3 The Definite Integral

465

The interval [a, b] can be deconstructed into n arbitrary subintervals [xi ; xi+1 ] with the (variable) width ∆ xi = xi+1 xi , i = 1, . . . , n. The area below and above the function is divided into (equal width) rectangles, whose heights are tangent to the function f (x) once on the left and once on the right. The surface area below f (x) in the interval [a, b] then clearly lies between the sum of all surfaces of all rectangles above and the sum of all surfaces of all rectangles below f (x).

11 Integral Calculus

466

To determine the area below the function f (x) in the interval [2; 5], the interval is, for example, divided into three equal rectangles whose heights touch the graph on the left. The sum of the areas of these rectangles is: 4 · 1 + 5 · 1 + 6 · 1 = 15 LU2 (LU = length units). Then the area is again divided into three rectangles of equal width, but their heights touch the graph on the right (equal abscissa intervals). Their area is: 5 · 1 + 6 · 1 + 7 · 1 = 18 LU2 . The area F, which is being searched for, is between the sum of the areas of the first rectangles and the sum of the areas of the second rectangles: 15 LU2 < F < 18 LU2 If the problem, or rather the procedure, is transferred to an arbitrary, continuous function f (x) over the interval [a, b], the following applies: n

n

i=1

i=1

∑ f (xi ) · ∆ xi ≤ F ≤ ∑ f (xi+1 ) · ∆ xi with ∆ xi = xi+1 − xi

i = 1, . . . , n

This approximation becomes more accurate the smaller the width of the intervals ∆ xi is. For the borderline case, where the width of the interval ∆ xi converges towards zero (∆ xi → 0), f (xi+1 ) strives towards f (xi ). The height of the rectangles below and above the graph of f (x) are then nearly identical. The sum of all areas of the rectangles above the function converges to the sum of all areas of the rectangles below the function, so that the desired area F with ∆ xi → 0 becomes more and more unambiguously in borderline cases unambiguously - determinable.

∆ xi → 0 with ∆ xi = xi+1 − x

The width of the formed rectangles below and above the function f (x) become smaller and smaller; the difference between the two areas converges to zero.

467

11.3 The Definite Integral

The smaller ∆ xi is selected, i.e. the more intervals are formed, the clearer the wanted area F can be determined. It can be determined (in borderline cases) if ∆ xi → 0 or n → ∞. n

n

F = lim ∑ f (xi )∆ xi = lim ∑ f (xi+1 )∆ xi = ∆ xi →0 n→∞

∆ xi →0 n→∞

i=1

i=1

Rb a

f (x)dx

The areas above and below the function f (x) practically coincide. The above mentioned limiting value of a function f (x), which is continuous in the interval [a, b], is called definite integral of the function f (x) in the limits a and b.

Remark: • The definite integral

Rb

f (x)dx is not a function but a fixed number.

a

The value of the definite integral can also be negative. • The definition of the definite integral can also be applied to discontinuous functions. So e.g. every piecewise continuous function with a finite number of jump discontinuities x1 , . . . , xn can be integrated. The integral

Rb a

sections.

x is the sum of the integrals over the single function

11 Integral Calculus

468

11.3.2 Relationship between the Definite and the Indefinite Integral The value of the definite integral is equal to the difference of the values of the antiderivative of the integrand f (x), F(x); value of the upper limit of the antiderivative F(x), F(b), minus value of the lower limit, F(a). Fundamental theorem of differential and integral calculus: Rb a

f (x)dx = [F(x)]ba = F(b) − F(a)

Examples:

(1)

Determination of the area below the function f (x) = x2 between x1 = 1 and x2 = 3: R3 2 x dx = 1



1 3 x 3

3

 =

1

   1 3 1 3 27 1 26 3 − 1 = − = 3 3 3 3 3

2 = 8 AU 3 AU

= area units = LU2

469

11.3 The Definite Integral

(2)

√

lower/upper limit: xl = 1, xu = 4 " #4   R4 1 R4 √ 1 3 2 3 4 2 3 2 3 14 xdx = x 2 dx = 3 x 2 = x2 = 42 − 12 = 3 3 3 3 1 1 1 2 f (x) =

x

1

2 = 4 AU 3

Variation of the Upper Limit If the lower integration limit a is kept constant and only the upper limit b is varied, there is exactly one area value F with F =

Rb

f (x) dx for each

a

value of the upper limit b. That means, there is a clear relation between f and b. To clarify this unique relationship, b is usually replaced by the independent variable x and the previous integration variable x is combined with another letter, for example t. Thus the value F of the integral from a to the upper (variable) limit x is written as: F = F(x ; t) =

Rx a

f (t)dt

t ∈ [a , x]

a, x ≥ 0

11 Integral Calculus

470

The function F(x) is called integral function of f (t) in the interval/area [a, x]. Example: f (t) = t Rx a

tdt = [F(t)]xa = F(x) − F(a) =

1 2 1 2 x − a 2 2

with a = const.

Depending on the definition of the lower integration limit a, the following integral functions are obtained: a = 0:

Rx

a = 2:

Rx

tdt =

1 2 1 2 1 x − · 0 = x2 2 2 2

tdt =

1 2 1 2 1 x − · 2 = x2 − 2 2 2 2

0

0

a = 10:

Rx 0

tdt =

1 1 2 1 x − · 102 = x2 − 50 2 2 2

Remarks: • The different integral functions of the example shown last merely differ by an additive constant. • During the formation of the definite integral, areas, which are above the x-axis, are valued positive and those, which are below the xaxis, are valued negative, thus on balance a value of zero or less than zero can result.

471

11.3 The Definite Integral Addition of the Absolute Values ⇒

Rb a

x R1 Rb f (x)dx = f (x)dx + · · · + f (x)dx a x n

with xi = zeros of the function f (x), i = 1, . . . , n

Example:

Solving the integral

Rb

f (x)dx is done by adding the absolute values of

a

the corresponding single areas A j , with j = 1, . . . , 5.

11 Integral Calculus

472 Example:

Remark: • The surface area of the piece of area, which is located between two function graphs f and g (with f ≥ g), is calculated as the difference between the two pieces of area located below the graphs: Rb F(x) = ( f (x) − g(x))dx a

In case f (x) and g(x) intersect within [a, b] with the points of intersection x1 , x2 , . . . , xn , the total area enclosed by the functions must be integrated from intersection point to intersection point to determine the total area enclosed by the functions: x R1 Rb F(x) = ( f (x) − g(x))dx + · · · + ( f (x) − g(x))dx a x n

with xi = point of intersection between the areas f and g, i = 1, . . . , n.

473

11.3 The Definite Integral

To avoid negative area dimensions, the absolute values are used again.

11.3.3 Special Techniques of Integration Unlike in differential calculus, integration rules do not exist for all integrable functions, i.e. there is no kind of "product rule", "quotient rule" or "chain rule". Instead, it is attempted to transform the integrand by suitable conversions into a form, which can be integrated in closed form by using basic integrals.

11.3.3.1 Partial Integration If the integrand is given as a product, the integral can often be transferred into a simpler form: R

f (x) · g′ (x)dx = f (x) · g(x) − f ′ (x) · g(x)dx R

with f , f ′ , g, g′ = continuous functions

11 Integral Calculus

474

This integration technique is related to the product rule of differential calculus: h(x) = f (x) · g(x) ⇒ h′ (x) = f ′ (x) · g(x) + f (x) · g′ (x)

Example of an indefinite integral: Find the solution for: lnx · xdx with D f = R+ R

1 g′ (x) = x ⇒ f ′ (x) = x R 1 2 R 1 1 2 ⇒ lnx · x dx = lnx · x − · x dx 2 x 2 2 R 1 x = lnx · − xdx 2 2   1 1 2 x2 = lnx · − · x +c 2 2 2

⇒ f (x) = lnx

= lnx ·

g(x) =

x2 1 2 − x −c 2 4

Example of a definite integral: R3

Find the solution for: x · ex dx 2

⇒ f (x) = x

g′ (x) = ex

⇒ f ′ (x) = 1

R3

R3

2

2

⇒ x · ex dx = [x · ex ]32 − 1 · ex dx = [x · ex − ex ]32 = [(x − 1)ex ]32

= (3 − 1)e3 − (2 − 1)e2 = 2e3 − e2 ≈ 32.78 AU

g(x) = ex

1 2 x 2

475

11.3 The Definite Integral 11.3.3.2 Integration by Substitution R

When integrating by substitution, the variable x in f (x)dx is replaced by a suitable function g(z). Provided that g(z) is differentiable and reversible, the following is valid: R

f (x)dx =

R

f (g(z)) · g′ (z)dz with x = g(z)

Example of an indefinite integral: R √ Find the solution for: x 1 − x2 dx

⇒ Substitution: 1 − x2 = z ⇒ dz = −2x dx or dx = −

1 dz 2x

R √ 1x R √ 1R 1 ⇒ x 1 − x2 dx = − z dz = − z 2 dz 2x 2 1 1 3 = − · 3 ·z2 +c 2 2 1√ = − 2 z3 + c 3 R √ 1p ⇒ Resubstitution: x 1 − x2 dx = − (1 − x2 )3 + c 3

Example of a definite integral: R2 √ Find the solution for: x3 x4 − 1dx 1

⇒ Substitution: z = x4 − 1 ⇒ dz = 4x3 dx

or dx =

1 dz 4x3

The original transformation limits xl = 1 and xu = 2 transform accordingly: zl = g(xl ) = 14 − 1 = 0 zu = g(xu ) = 24 − 1 = 15

11 Integral Calculus

476 R15 1 √ R2 √ R15 √ 1 ⇒ x3 x4 − 1dx = x3 z · 3 dz = zdz 4x 1 0 0 4 " #15 R15 1 1 1 1 3 = z 2 dz = · ·z2 4 32 0 4 0

=

1 3 1 3 · 15 2 − · 0 2 ≈ 9.68 AU 6 6 "

1 1 3 Resubstitution: · ·z2 4 32 =

#15 0




3 1 4 = x −1 2 6

2 1


3 1
3 1 4 2 − 1 2 − 14 − 1 2 ≈ 9.68 AU 6 6

11.4 Multiple Integrals A function with several independent variables f = f (x1 , . . . , xn ) can be integrated by partially integrating c.p. (= if the remaining variables are constant) successively after all variables: R

···

RR

f (x1 , x2 , . . . , xn ) dx1 dx2 . . . dxn .

Remark: The innermost integral symbol belongs to dx1 , the next following symbol to dx2 and the outer symbol to dxn . The integration is done from inside to outside.

Example of an indefinite double integral:  RR

xy dx dy =

R

 1 2 1 x y + c(y) dy = x2 y2 +C(y) + d(x) 2 4

11.5 Integral Calculus and Economic Problems

477

Example of a definite double integral: R5 R3

1 dx dy =

2 1

R5 2


R5 R5 [x]31 dy = (3 − 1)dy = 2dy 2

2

= [2y]52 = 10 − 4 = 6 AU (here in three-dimensional space)

11.5 Integral Calculus and Economic Problems The relationship between total economic functions and marginal economic functions is illustrated by means of the definite integral. Remark: By definition, total economic functions are always antiderivatives of the corresponding marginal economic functions.

11.5.1 Cost Functions Let C′ (x) be the marginal cost function of the total cost function C(x). ⇒

Rx ′ C (q)dq = C(x) +C f 0

⇒

or

Rx ′ C (q)dq = Cv (x) 0

Rx C(x) = C′ (q)dq +C f

or C(x) = Cv (x) +C f

0

with Cv (x) = variable costs; C f = fixed costs The integral of the marginal function thus corresponds to the variable costs Cv (x). The following relationships apply between the total costs C(x), the marginal costs C′ (x), the variable costs Cv (x) and the fixed costs C f :

11 Integral Calculus

478 Rx

Cv (x) = C′ (q)dq

or

0

Rx

C(x) = C′ (q)dq +C f 0

Graphically, the variable costs Cv (x) for the output x correspond to surface area of the area below the marginal costs between zero and x.

Example: The fixed costs C f of $4, 000 and the marginal cost function C′ (x) = 0.03x2 − 3x + 120 [$/unit] are known. What are the total costs with an output x of 400 QU (quantity units)?

11.5 Integral Calculus and Economic Problems

479

C(x) = Cv (x) +C f 400 R


0.03x2 − 3x + 120 dx + 4, 000 0  400 1 3 1 2 = 0.03 · · x − 3 · · x + 120x + 4, 000 3 2 0   = 0.01 · (400)3 − 1.5 · 4002 + 120 · 400 − (0) + 4, 000 =

= $452, 000

11.5.2 Revenue Function (= Sales Function) R′ (x) is the marginal revenue function of the revenue function R(x).

R(x) =

Rx ′ R (q)dq 0

Graphically, the total revenue R(x) for the quantity sold x corresponds to the surface area underneath the curve of the marginal revenue between 0 and x. Note: The areas located below the x-axis are negative.

11 Integral Calculus

480 Example:

The marginal revenue function is R′ (x) = 1, 044 − 0.6x [$/unit]. What is the revenue function and what is the associated inverse demand function (demand function)? Revenue function:

R(x) =

Rx

(1, 044 − 0.6q) dq

0

1 = 1, 044q − 0.6 · q2 2 1, 044x − 0.3x2

=

x 0

[$]

Inverse demand function: R(x) = x · p(x) ⇔ p(x) =

R(x) = 1, 044 − 0.3x [$/unit] x

11.5.3 Profit Functions The (total) profit P(x) is determined by the difference between revenue R(x) and total costs C(x), so that: 

Rx Rx ′ P(x) = R(x) −C(x) = R′ (q)dq − C (q)dq +C f 0

=

Rx



0

(R′ (q) −C′ (q)) dq −C f

0

This results in the contribution margin PCM (x) : PCM (x) =

Rx

(R′ (q) −C′ (q)) dq

0

Graphically the contribution margin PCM (x) is obtained for the sold quantity x as a measure of the area between the marginal revenue and the marginal cost curve.

11.5 Integral Calculus and Economic Problems

481

Note: If R′ is below C′ , the surface pieces are evaluated as negative, so that the total contribution margin is the difference of the positive and negative evaluated areas.

PCM (x) = contribution margin for quantity x

Example: The marginal cost function C′ (x) = 3x2 − 24x + 60 as well as the marginal revenue function R′ (x) = −18x + 132 are given. The total costs for the output of 10 QU (quantity units) amount to $498.

Determine the following: (1) the revenue function, (2) the total cost function, (3) the inverse demand function, (4) the profit function.

11 Integral Calculus

482

for (1): R(x) =

Rx ′ Rx R (q)dq = (−18q + 132) dq 0

0

x  1 = −18 · · q2 + 132q 2 0 = −9x2 − 132x [$]

Rx

Rx

0

0

for (2): C(x) = C′ (q)dq + K f =


3q2 − 24q + 60 dq +C f

x  1 1 = 3 · · q3 − 24 · · q2 + 60q +C f 3 2 0 = x3 − 12x2 + 60x +C f [$] ⇒ C(10) = $498 ⇒ 103 − 12 · 102 + 60 · 10 +C f = $498 ⇒ C f = 498 − 400 = $98 ⇒ C(x) = x3 − 12x2 + 60x + 98 [$]

for (3): R(x) = x · p(x) ⇒ p(x) =

R(x) −9x2 + 132x = x x

= −9x + 132 [$/unit]

483

11.5 Integral Calculus and Economic Problems x for (4): P(x) = R(x) −C(x) =

R

(R′ (q) −C′ (q)) dq −C f



0

⇒ P(x) = −9x2 + 132x − x3 − 12x2 + 60x + 98 = −x3 + 3x2 + 72x − 98 [$]




Chapter 12

Elasticities 12.1 Definition of Elasticity The subject of this chapter is the analysis of the relative rate of change of economic variables when there is a functional relationship between them, for example y = y(x).

Absolute Changes ⇒ Difference quotient

∆ y(x) ∆x

= average absolute slope of the function y(x) in a specific interval ⇒ Differential quotient

d f (x) (x0 ) = 1st derivative at the point x0 dx

= slope of the function y(x) at any point, at any position x0 that relates to an infinitesimal area around x0 Interpretation/Question: By how many units does the dependent variable y change when the independent variable x varies by 1 unit?

Relative Changes By what percentage does the dependent variable change if the independent variable varies by 1 %? (1)

related to a certain interval ⇒ arc elasticity

(2)

related to a certain point (at a certain location) ⇒ point elasticity

© Springer-Verlag GmbH Germany, part of Springer Nature 2023 F. W. Peren, Math for Business and Economics, https://doi.org/10.1007/978-3-662-66975-4_12

485

12 Elasticities

486

12.2 Arc Elasticity The given function is y = y(x). The ratio of the relative (= percentage) changes is called arc elasticity εA (= the average elasticity) of y with reference to x: ∆y relative change of y y εA = = ∆x relative change of x x

with

y = y(x)

⇒ the relative (= percentage) change of the dependent variable is put into relation to the relative (= percentage) change of the independent variable. εA is dimensionless. Example: The inverse demand function is: p = p(x) = 20 − 0.2x or x = x(p) = 100 − 5p

(demand function = inverse function solved for x)

12.2 Arc Elasticity

487

General question:

By what percentage does the dependent variable (here the quantity demanded of a good x) change, on average if the independent variable (here the price of good p) changes by one percent ?



 ∆x  relative quantity change x   ⇒  relative price change = ∆ p  p The quotient is called arc elasticity (in the considered interval/arc ∆ p). Examples of absolute changes: Case 1

Case 2

Previous price: p

15

2

Price change: ∆ p

−1

−1

⇒ new price: p + ∆ p

14

1

Previous quantity: x

25

90

Quantity change: ∆ x

+5

+5

⇒ new quantity: x + ∆ x

30

95

⇒ In both cases, a price reduction of $1 per unit determines an absolute change in demand by +5 QU. = 1st derivative:

dx = −5 dp

The (absolute) slope of the demand function is the same for all x. If the price is reduced by $1 per unit, the quantity demanded is (constantly) reduced by 5 QU. An absolute price reduction of for example $15 to $14 by $1 per unit is assessed relatively differently (−6.67 %) than the same absolute change about $1 per unit from e.g. $2 to $1 per unit (−50 %).

12 Elasticities

488 Examples of absolute changes:

price change ∆ p  =

quantity change

Case 2

−6.67 %

−50 %

 14 − 1 · 100 % 15

 =

+20 %  =

arc elasticity εA

Case 1

 30 − 1 · 100 % 25

 1 − 1 · 100 % 2

+5.56 %  =

 95 − 1 · 100 % 90

+20 % −6.67 %

+5.56 % −50 %

≈ −3

≈ −0.11

12.2 Arc Elasticity

489

Example: ∆x x(p + ∆ p) − x(p) 30 − 25 x(p) x(p) 25 εA = = = = −3 14 − 15 ∆p (p + ∆ p) − p 15 p p

Interpretation of εA = 3 : If the price of the good p rises or falls by 1 %, demand x falls or rises by an average of 3 % in the price range between $14 and $15 per unit.

12 Elasticities

490 Example: y(x) = x2 + 1 εA between x1 = 3 and x2 = 4?

x1 = 3 ⇒

∧

x2 = 4

⇒ ∆ x = +1

y(x1 ) = y(3) = 32 + 1 = 10 y(x2 ) = y(4) = 42 + 1 = 17

⇒

∆ y = +7

+7 ∆y (= +70%) x ≈ 2.1 ⇒ εA = = 10 +1 ∆x (= +33%) 3 x

i.e.: If the independent variable x increases or decreases by 1 %, the value of the function y rises or falls on average by about 2.1 % in the interval between x1 = 3 to x2 = 4.

12.3 Point Elasticity

491

12.3 Point Elasticity While the arc elasticity indicates the average rate of change within an interval, in economic sciences it is usually the elasticity around a certain point that is of interest, i.e. at a certain place x0 . ⇒ determination of the limit value of the arc elasticity for ∆ x → 0

⇒

∆y ∆y x y lim εA = lim · = = lim ∆ x ∆ x→0 ∆ x→0 ∆ x→0 y ∆x x ∆y x dy x x · = · = y′ (x) · = ε ∆ x→0 ∆ x y dx y y

= lim

ε(x) = (point) elasticity function ε(x = x0 ) = (point) elasticity or, in other words, elasticity of the function y = y(x) at the point x = x0 . ε is dimensionless.

Definition y is a differentiable function with the independent variable x. This means then dy dy x x y ε(x = x0 ) = = · = y′ (x) · dx dx y y x

with

x, y ̸= 0

is called (point) elasticity ε of the function y = y(x) at the point x = x0 . The numerical value of the (point) elasticity ε of y related to x at a specific point x = x0 indicates (approximately) by what percentage the dependent variable y changes if the independent variable x (at this point x0 ) varies marginally by 1 %; with y = y(x).

492

12 Elasticities

Note: The sign of elasticity ε plays an important role (Tab. 12.1): dy y (1) If ε > 0 with y = y(x) then > 0 applies by definition, dx x i.e. the relative changes of the considered variables are either both positive or both negative. Thus a relative increase (decrease) of x causes a relative increase (decrease) of y. y and x are positively correlated. (2) If ε < 0, a relative increase (decrease) of x causes a relative decrease (increase) of y. y and x are negatively correlated. (3) If ε = 0, y remains constant with an increase (decrease) of x. y and x are not correlated. Example: f (x) = x2 − x + 10 What is the (point) elasticity of f at x0 = 10?

ε = ε(x) =

=

ε(10) =

d f (x) d f (x) x x f (x) = · = f ′ (x) · = dx dx f (x) f (x) x (2x − 1) · x 2x2 − x = = elasticity function x2 − x + 10 x2 − x + 10

2 · 102 − 10 = 1.9 = (point) elasticity at x = 10 102 − 10 + 10

Interpretation: If x is increased (decreased) by 1 % at the position x = 10, the value of the function f (10) will then be increased (decreased) by 1.9 % (disproportionately). The relation between x and f (x) is “elastic”. f (x) and x are positively correlated.

12.3 Point Elasticity

493

Value of Elasticity

General Definition Example:

| εxp |< 1

x is inelastic

Demand Function x = x(p) with p = price and x = quantity Relatively low reaction of the consumer to price changes.

0 1 or ε > −1

(x changes relatively Example: substitutable goods stronger than p)

Special case: | εxp |= 1

x is proportionally elastic; isoelastic

ε = 1 or ε = −1

(the relative changes of x and p are equal)

Borderline case: | εxp |→ ∞

x is perfectly elastic Borderline case: The price is constant, regardless of the level of demand.

ε → ∞ or ε → −∞

(x reacts infinitely strong to small relative changes of p)

Relatively strong reaction of the consumers to (small) relative price changes.

A price change of 1 % causes a proportional quantity change of 1 %.

Example: fixed-price substitutable goods (branded articles in the polypoly).

12 Elasticities

494

Borderline case: ε=0

x is perfectly inelastic; rigid

Borderline case: The demand is constant, i.e. independent of the price.

(x does not react to Example: indispensable goods insignificant relative such as essential medicines. changes of p) Tab. 12.1: Elasticities | Case Distinction

12.4 Price Elasticity of Demand εxp Definition The price elasticity of demand εxp measures the relative change in demand in consequence of a relative change of the price by 1 % at a specific point (x0 |p0 ).

εxp

dx dx(p) p = x = · dp dp x p

Attention: • dependent varibale

=

quantity demanded x

• independent variable

=

price p

⇒ inverse demand function: x = x(p)

Note: If the inverse demand function p = p(x) is used, the price elasticity of demand εxp would be required analogously.

12.4 Price Elasticity of Demand εxp

495

Example: Given, the inverse demand function is p(x) = 10 − 0.5x. Searched is the price elasticity of demand at a price of p0 = $6/QU. Since in this case, p is the independent variable and x the dependent variable, the demand function x = x(p) must be formed. ⇒ x(p) =

εxp

10 − p = 20 − 2p 0.5

dx p dx(p) p · = x′ p(x) · = x = dp dp x x(p) p

x(p)

= 20 − 2p

x′ (p)

= −2

p0

= $6/QU

x(6)

= 20 − 2 · 6 = 8 QU

εxp

= −2 ·

6 = −1.5 8

Interpretation: With regard to the basic price p0 = $6/QU, an increase (decrease) in price by 1 % causes a demand decrease (increase) by approximately 1.5 %. Consumers react elastically at a price of p0 = $6/QU (| εxp |> 1).

12 Elasticities

496 Example:

Given, the inverse demand function is p(x) = 10 − 0.5x. The price or quantity intervals are to be found, where the demand is (a) elastic, (b) inelastic, (c) proportionally elastic (isoelastic), (d) perfectly inelastic, (e) perfectly elastic.

εxp

dx dx(p) p = x = · dp dp x p

⇒ x(p) =? p = 10 − 0.5x ⇔ 0.5x = 10 − p

⇔

x(p) = 20 − 2p

(a) The demand is price elastic if the following applies: εxp < −1. Since the slope of the inverse demand function is negative (quantity and price are here negatively correlated), the case of elasticity is not applicable. Hence, εxp > 1. εxp = x′ (p) ·

p −p p = −2 · = < −1 x(p) 20 − 2p 10 − p

⇔ −p < −1(10 − p) ⇔ −p < −10 + p ⇔ −2p < −10 ⇔ 2p > 10 ⇔

p > $5/QU

12.4 Price Elasticity of Demand εxp

497

Interpretation: The demand is price elastic for prices between $5 and $10 per unit. The corresponding quantity range is between 0 and 10 units. (b) The demand is price inelastic, if: εxp > −1 εxp =

−p > −1 10 − p

⇔ −p > −1(10 − p) ⇔ −p > −10 + p ⇔ −2p > −10 ⇔ 2p < 10 ⇔

p < $5/QU

Interpretation: The demand is price inelastic for prices between $0 and $5 per unit. The corresponding quantity interval is between 10 and 20 units. (c) The demand is isoelastic, if: εxp = −1 −p = −1 10 − p ⇔ p = $5/QU εxp =

⇒ x = 10 QU

Interpretation: The demand is isoelastic if the price is $5 per unit. If that price changes by 1 % ($4.95 or $5.05), a proportional quantity change of 1 % (9.90 QU or 10.10 QU) is affected.

12 Elasticities

498

(d) The demand is perfectly price inelastic, if: εxp = 0 εxp =

−p =0 10 − p

⇔ p → $0/QU ⇒ x = 20 QU

Interpretation: If the price converges towards zero or even becomes zero, all 20 QU are sold. Consumers do not react at all to (marginal) relative price changes. (e) The demand is perfectly price elastic, if: εxp = ∞ εxp =

or

εxp = −∞

−p → −∞ 10 − p

⇔ p → $10/QU ⇒ x = 0 QU

Interpretation: If the price converges towards $10/QU or even becomes $10/QU, nothing can be sold. Consumers react perfectly elastic to (marginal) relative price changes.

12.5 Cross Elasticity of Demand εxA pB

499

12.5 Cross Elasticity of Demand εxA pB The quantity demanded xA of a good A depends not only on the (relative) changes of its own price pA but also on the price changes of other goods, such as good B: xA = xA (pA , pB ). This relationship is known as the cross elasticity of demand (or crossprice elasticity of demand), εxA pB : εxA pB =

relative change in quantity of good A relative change in price of good B

Related to an infinitesimal range at a specific point xA = xA (pA , pB ), the following applies:

εxA pB

∂ xA ∂ xA pB x = A = · ∂ pB ∂ pB xA (pA , pB ) pB

12 Elasticities

500 Example:

The demand xα for the type of notebook “Alpha” depends on the price pα of its own system and the price pβ of the competitors’ system “Beta”. The corresponding demand function is: xα = 10, 000 − 2pα + 3pβ The system prices are currently at: pα = $2, 000 per unit α

and

pβ = $2, 200 per unit β

What is the cross elasticity of α in relation to the price of β at the current price situation?

εxα pβ

∂ xα pβ ∂ xα x = α = · = ∂ pβ ∂ pβ xα (pα , pβ ) pβ pβ 3 · 2, 200 = 3· = ≈ 0.5238 10, 000 − 2pα + 3pβ 10, 000 − 2 · 2, 000 + 3 · 2, 200

Interpretation: A price increase (decrease) of the system “Beta” by 1 % causes a demand increase (decrease) for “Alpha” by 0.52%. xα and pβ are positively correlated, however in an inelastic case. There is no significant substitution effect.

12.6 Income Elasticity of Demand εxy

501

12.6 Income Elasticity of Demand εxy The quantity demanded x of a good depends on the changes of the consumers’ income y: x = x(y). This relationship (regarding relative changes of the relevant variables) is known as the income elasticity of demand εxy .

εxy

dx dx y y = x = · = x′ (y) · dy dy x(y) x(y) y

Example: The saleable quantity x of a certain vehicle depends on the net monthly income y in relation to a certain region. The consumption function is: x = 5, 500 + 50y Last year, the total sales in that region were 100, 500 vehicles and the average net monthly income was $2, 600. What was the income elasticity of demand?

εxy = x′ (y) ·

50 · 2, 600 y = ≈ +0.96 x(y) 5, 500 + 50 · 2, 600

Interpretation: An increase of the net income by 1 % (= $26) results in an increase in demand of that vehicle by 0.96 %. This indicates an isoelastic case.

Chapter 13

Economic Functions 13.1 Supply Function The supply function represents the relationship between the market price of a good (independent variable) and the quantity supplied (dependent variable) in the form of a unique graph (function). Typically, a high price indicates a large quantity of goods supplied. If the price falls, the quantity offered is also usually reduced. This is depicted by the supply function x = x(p) with x = quantity supplied and p = offer price. An analogous explanation can be given for the inverse function p = p(x), the inverse supply function (Fig. 13.1). A large supply exists at a high, realizable price. The supply is small when the prices of this tradable good or service offered are low and thus unattractive for the supplier. In macroeconomics, a distinction is made between an individual supply function and the aggregate of individual supplies. If the state of perfect competition exists, the marginal costs correspond to the offer price and the marginal cost function coincides with the supply function. The supply function runs strictly monotonically increasing. It can be linear, as shown in the figure, or partially or completely curved (concave or convex). If the supply function is flat, the price elasticity of the supply or price elasticity of the suppliers is expected to be high. Relative price changes then cause disproportionately high, relative quantity changes. This means that the reaction of suppliers to price changes is disproportionately strong. A (tendentially) steep course of the supply function indicates a relatively low supply elasticity.

© Springer-Verlag GmbH Germany, part of Springer Nature 2023 F. W. Peren, Math for Business and Economics, https://doi.org/10.1007/978-3-662-66975-4_13

503

13 Economic Functions

504

Fig. 13.1: Inverse Supply Function

Example 1: The (inverse) supply function, which reflects the behaviour of suppliers in relation to price, is: x(p) = 4p − 10. If one wants to estimate the quantity supplied at a currently prevailing price of $20, the following calculation is to be made: x(20) = 4 · 20 − 10 = 70 QU

QU

= quantity unit(s)

With an offer price of $20, the corresponding quantity supplied that is offered is 70 QU. Example 2: Suppliers’ behaviour of a manufacturing firm can be described using the supply function p(x) = 12.5x + 4. The corresponding offer price for a quantity supplied of 5 QU shall be calculated as follows: p(5) = 12.5 · 5 + 4 = $66.5 With a quantity supplied of 5 QU, the offer price is $66.5.

13.2 Demand Function / Inverse Demand Function

505

13.2 Demand Function / Inverse Demand Function The demand function x(p) represents the quantity demanded x (QU) for a good or a service depending on the market price p ($/QU = $ per unit): x = x(p) with x = quantity demanded and p = asking price The inverse demand function p(x) is the inverse function of a demand function: p(x) = f −1 (x(p)). p(x) views the price p as a function of quantity x. The inverse demand function p(x) treats the price as a function of quantity demanded. It is also called the price function. In contrast to the supply function, the inverse demand function (Fig. 13.2) is usually strictly monotonically decreasing. If the market price rises/falls, the quantity demanded falls/increases.

Fig. 13.2: Inverse Demand Function For certain goods or services, e.g. luxury goods or services, the relationship between demanded quantity and price can also be reversed, i.e. when the price rises/falls, quantity demanded rises/falls (Giffen’s paradox, snob effect). Similar to the supply function, a distinction is made between an individual and an aggregated demand function.

506

13 Economic Functions

The graphical representation of the inverse demand function is marked by two significant points: the saturation quantity and the prohibitive price. The saturation quantity is determined at a price of $0 per unit; in the graph, the saturation quantity corresponds to the intersection of the inverse demand function with the x-axis. This corresponds to the highest possible quantity demanded for a good or service. Analogously, the prohibitive price is determined by setting the quantity demanded to zero. This state occurs when the price is so high that no one demands this good or service. Graphically, the amount of the prohibitive price is determined by the y-intersect of the inverse demand function.

Example 1: The demand function for a manufacturer of coffee cups is x(p) = −2p + 7. If the price per cup is set at $3, the quantity demanded is determined as follows: x(3) = −2 · 3 + 7 = 1 QU One coffee cup is demanded at $3.

Example 2: If the prohibitive price of the inverse demand function is p(x) = −0.2x + 10, the quantity demanded x is equated to zero and the following is obtained: p(0) = −0.2 · 0 + 10 = $10/QU At a price of $10, no quantity is demanded.

13.3 Market Equilibrium

507

Example 3: For the inverse demand function p(x) = −0.25x + 17, the direction of causality is changed. p(x) = −0.25x + 17 must be calculated in terms of x as follows: p − 17 = −0.25x p − 17 =x −0.25 −4p + 68 = x x = x(p) = −4p + 68

13.3 Market Equilibrium When the supply and demand of a good or service in the market coincide, the market is in equilibrium. In the market equilibrium (Fig. 13.3) the quantity supplied is completely demanded or sold on the market. The market is “cleared”. Therefore, it is also called market clearing. The equilibrium price and the equilibrium quantity represent the perfect market. The quantity supplied corresponds exactly to the demanded quantities of the consumers. There is no waste of this good. Also, a further exchange would not be an improved alternative. In reality, this state is often not reached due to insufficient transparency. If goods or services are (also) traded online, transparency is usually improved, which in turn should bring trade closer to market equilibrium. Graphically, the market equilibrium is described by the intersection of the supply and inverse demand function.

508

13 Economic Functions

Fig. 13.3: Market Equilibrium

13.4 Buyer’s Market and Seller’s Market The position in which a buyer or seller is situated determines the market situation in comparison to the market equilibrium. If supply is greater than demand at a given price, the buyer is in a relatively better position than the seller. The market surplus defines the market as the buyer’s market. A market surplus exists when the supply of a good or service exceeds the demand. The supplier(s) can manage this imbalance to their benefit by reducing their supply and minimising their surplus. Likewise, a market shortage leads to a seller’s market. If the demand is greater than the supply, this strengthens the position of the seller in relation to the buyer. In a monopolistic market, higher prices can usually be set so that the seller can dominate the market. Also, a densification of the market as a result of mergers can lead to the fact that substitute goods are rarely or never found on the market. Even in the case of emergency goods, such as medicines, a certain (temporary) dependence of the buyer on the seller can arise and shift the price away from the market equilibrium at the expense of the buyer. Other reasons for the occurrence of a seller’s market can be an overregulated market, paucity of information of the buyer or a lack of competition.

13.6 Demand Gap

509

13.5 Supply Gap A supply gap exists if the quantity supplied is smaller than the quantity demanded for a particular good or service in a specific market.

13.6 Demand Gap Unlike the supply gap, the demand gap is a phenomenon where the quantity supplied is greater than the quantity demanded for a good or service in a specific market. Both the supply gap and the demand gap refer to an imperfect market condition. Example 1: The market equilibrium between supply and demand should be identified. The supply can be described by the supply function p(x) = 0.5x + 9 and the demand by the inverse demand function p(x) = −0.75x + 13. Thus, the point is being searched, i.e. the quantity and the price, where the supply function and the inverse demand function intersect: supply function = inverse demand function 0.5x + 9 = −0.75x + 13 −4 = −1.25x x = 3.2 QU p(3.2) = 0.5 · 3.2 + 9 = $10.6/QU The market equilibrium is located at the point (3.2|10.6).

13 Economic Functions

510 Example 2:

Supposedly, the inverse demand function shifts due to an increase in income. The question to be discussed is how the initial equilibrium consequently develops. The inverse demand function is shifted upwards to the right in response to the increased income because at any given
price p, a larger quantity x can now be demanded. A new point p(x0 )|x0 arises which describes the new market equilibrium. Example 3: To protect a firm’s production, a lower price limit of $5/QU is implemented. With the inverse demand function pD (x) = −6x + 17 and the supply function pS (x) = 2x − 3, it is to be determined whether a price of $5/QU leads to a surplus/a demand gap or to a shortage/a supply gap. The fixed price $5/QU can be placed in both functions:

Inverse demand function:

Supply function:

5 = −6x + 17

→ xD = 2 QU

5 = 2x − 3

→ xS = 4 QU

With a lower price limit of $5/QU, the supply is greater than the demand. In this case, there is a surplus/a demand gap of 2 QU.

13.7 Revenue Function

511

13.7 Revenue Function The revenue function describes the behaviour of the revenue R in relation to the quantity sold x. The quantity sold x depends on the price p, which can either be constant or vary depending on the quantity p = p(x). If the inverse demand function p(x) is multiplied by the quantity x, the revenue function is obtained depending on the quantity x:

R(x) = p(x) · x

with p = selling price x = quantity purchased / sold

Similarly, the revenue function can also be modeled in relation to the price p:

R(p) = x(p) · p

with p = selling price x = quantity purchased / sold

Two different cases must be distinguished: 1. The price p is constant The quantity sold x does not change in response to price alterations (Fig. 13.4). It is not influenced by discounts or other price alterations. There is no causal relationship between the selling price p and quantity sold x. The revenue (R = R(x) = p · x) is proportional to the quantity. In other words, the revenue function is linear. The slope of the revenue function, i.e. the first derivative of R(x), corresponds to the constant price p. The intercept is zero. If no sales are generated (x = 0) the revenue is also zero. A relative maximum in revenue does not exist because the revenue grows constantly, i.e. linearly, along with the quantity sold x, so that the maximum is at the right edge of the revenue function, i.e. at x = xmax . The minimum in revenue is zero and is located at the origin of the revenue function at the point (0|0) where there are no sales generated, i.e. at x = xmin = 0. If the good or service is not sold, no revenue is generated, Rmin = R(0) = 0.

512

13 Economic Functions

Fig. 13.4: Behaviour of Revenues Based on a Linear Revenue Function

2. The price p = p(x) is variable The quantity sold x = x(p) changes due to price alterations p (Fig. 13.5). There is a causal relationship between the selling price p = p(x) and the quantity sold x = x(p). The revenue R = R(x) = p(x) · x is now generated according to the inverse demand function at variable prices with p(x) = −mx + b. This is modeled by a quadratic revenue function: R(x) = p(x) · x = (−mx + b) · x = −mx2 + bx The revenue function resembles a parabola opening downwards that starts at the origin (0|0) and ends at the point (xmax |0). xmax represents the saturation quantity, i.e. the maximal quantity sold at a price of pmin = 0.

513

13.7 Revenue Function The two zeros are calculated as follows: −mx2 + bx = 0 x(−mx + b) = 0 x = 0 ∨ −mx + b = 0 x=0 ∨ x=

b m

The relative maximum of the revenue function is calculated as follows: R(x) = −mx2 + bx dR(x)/dx = −2mx + b = 0 x=

b 2m

d 2 R(x)/dx2 = −2m < 0 → the maximum is at x =

b 2m

The maximum in revenue (maximum of the revenue function) is located at the point with the coordinates

x=

   2   b b b b −b2 b2 and R = −m +b = + . 2m 2m 2m 2m 4m 2m

The first zero is derived from a quantity sold of xmin = 0. The second b zero at xmax = describes the scenario where the price p = pmin = 0, m i.e. the saturation quantity is reached. The market is flooded with this good or service to such an extent that the price becomes zero. The quantity inflates until it reaches the price p = 0. The saturation quantity is the quantity that is reached at a price of 0. Half of the saturation quantity is the quantity sold that generates the maximum revenue.

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514

Fig. 13.5: Behaviour of Revenues Based on a Quadratic Revenue Function

Example 1: A manufacturer offers a product at a fixed price of $5/QU; p = $5/QU. In a given period, 300 units of this product are sold; x = 300 QU. The revenue of this company with an output quantity of 300 QU is therefore R(x) = $5/QU · 300 QU = $1, 500.

Example 2: An inverse demand function is given as p(x) = −2x + 20. In order to determine the revenue function, the inverse demand function must be multiplied by x. The revenue function is calculated as follows: R(x) = (−2x + 20) · x = −2x2 + 20x. The two zeros of the revenue function are calculated below: R(x) = 0 −2x2 + 20x = 0

515

13.7 Revenue Function x2 − 10x = 0 p x1 = − + 2 10 x1 = + 2

r  p 2 −q 2

s −

10 2

2 −0

√ x1 = 5 + 25 − 0 x1 = 5 + 5 = 10 The first zero is at (10|0). R(x) = 0 p x2 = − − 2 10 x2 = − 2

r  p 2 −q 2

s

10 − 2

2 −0

√ x2 = 5 − 25 − 0 x2 = 5 − 5 x2 = 0 The second zero is at (0|0). With the quantities produced of 0 QU and 10 QU, there is no revenue generated. Test: The two zeros of a revenue function that are dependent on the b price are x = 0 and x = (see above). In this case, the zeros are m

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516

20 = 10 QU, which corresponds to the solution 2 shown above. The saturation quantity is 10 QU. at x = 0 QU and at x =

To determine the maximum of the revenue function, the derivation of this function is required. The first derivative of the revenue function dR(x) R(x) = −2x2 + 20x is = −4x + 20. Following this, the marginal dx revenue function is set to zero: dR(x) =0 dx −4x + 20 = 0 4x = 20 x=5 d 2 R(x) = −4 < 0 dx2 The maximum of the revenue is located at x = 5 QU at a price of p(5) = −2 · 5 + 20 = $10/QU. The maximal revenue that can be achieved is R(5) = 10 · 5 = 50 QU. Test: The maximal revenue (peak of a revenue function which is depenb dent on the price) is located at the point with the coordinates x = 2m   b −b2 b2 and R = + (see above): (2m) 4m 2m 20 x= = 5 QU; (2 · 2)     b 20 −202 202 =R = R + = −50 + 100 = 50 QU. (2m) 4 4·2 2·2

13.8 Cost Functions

517

13.8 Cost Functions The total cost function shows the relationship between the total production costs within a particular process or for manufacturing a product C and the quantity produced or (externally) purchased x (Fig. 13.6). The total cost function usually contains a fixed cost portion, the fixed total costs C f and a variable cost portion, the variable total costs Cv (x): C(x) = C f +Cv (x) with x = produced or (externally) purchased

Fig. 13.6: Behaviour of the Total Costs Based on a Linear Cost Function

518

13 Economic Functions

The unit costs c(x) can be calculated with c(x) =

C(x) (Fig. 13.7). x

Fig. 13.7: Behaviour of the Unit Costs Based on a Linear Cost Function The fixed costs C f are independent of any business activities (Fig. 13.8), i.e. of production or quantity produced. The fixed costs remain unaffected by any change in the quantity produced. Fixed costs are, for example, warehouse rental costs.

Fig. 13.8: Behaviour of the Fixed Costs Based on a Linear Cost Function

13.8 Cost Functions

519

If the fixed costs are distributed linearly over the produced or purchased Cf quantity x, the result is the fixed unit costs . The fixed unit costs dex crease as the quantity increases (Fig. 13.9). There is a so-called fixed cost degression, since the fixed costs per unit decrease (degressively) with each unit that is additionally produced/purchased.

Fig. 13.9: Behaviour of the Fixed Unit Costs Based on a Linear Cost Function The variable costs Cv (x) cover the portion of the total cost function that varies with the quantity produced (Fig. 13.10). Examples of variable costs are commissions or the material or energy costs incurred for production. According to the costs-by-cause principle, variable costs can be assigned to every single unit that is produced/purchased in form Cv (x) of variable unit costs cv (x) with cv (x) = . For a linear cost function, x dC(x) the variable unit costs correspond to the marginal cost = cv (x) dx and are constant (Fig. 13.11).

520

13 Economic Functions

Fig. 13.10: Behaviour of the Variable Costs Based on a Linear Cost Function

Fig. 13.11: Behaviour of the Variable Unit Costs Based on a Linear Cost Function

521

13.8 Cost Functions

The marginal cost function is the first derivative of the total cost function: dC(x) = C′ (x) dx

.

The marginal cost function describes the slope of the total cost function in dependence of the quantity x (Fig. 13.12). The marginal cost x = x0 measures the expense incurred in producing or procuring one additional unit of that good, or the reduction in total costs that results from reducing the production or procurement of that good by one unit, x = x0 . In a linear cost function, the behaviour of the marginal costs corresponds exactly to that of the variable unit costs.

Fig. 13.12: Behaviour of the Marginal Costs Based on a Linear Cost Function

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13 Economic Functions

If the total cost function is not linear, it can also behave alternately, as shown graphically in Fig. 13.13. The mathematical principles of alternative cost trends are given in Tab. 13.1.

Fig. 13.13: Alternative Cost Trends • Degressive cost trend: As the production/procurement increases, the total cost C(x) has a disproportionately slow increase with a growing quantity x. For example, if the quantity produced increases by 5 %, the production costs increase by only 3 %. A degressive tendency of the total costs is known, for example, in the (continuous) offering of discounts. The procurement costs per unit (unit costs) decrease with the increasing quantity procured. • Progressive cost trend: As the production/procurement increases, the total cost C(x) has a disproportionately fast increase with a growing quantity x. For example, if the quantity produced increases by 5 %, the production costs increase by 7 %. Labour costs can rise progressively if, for instance, overtime is paid at a rate that increases disproportionately over time. The labour costs per unit (unit labour costs) then rise progressively with an increasing amount of overtime hours. • Regressive cost trend: As the production/procurement increases, the total cost C(x) has a disproportionately slow decrease with a growing quantity x.

523

13.8 Cost Functions

• Step-fixed cost trend: As the production/procurement increases, the fixed costs C f (x) included in the total costs C(x) grows rapidly (step-fixed) with an increasing quantity x at a certain position x = x0 . This can occur during the course of development to the same or a different extent. A sharp (step-fixed) decrease of the fixed costs at a certain position x = x0 is also possible.

Behaviour

C(x)

Example

Marginal costs dC(x)/dx

proportional

bx

x

b

b

degressive

x d with b < d

√

x

b b −1 xd d

x d −1

progressive

bxd

x2

dbxd−1

bxd−1

regressive

bx−d

1 x

(−d)bx−d−1

bx−d−1

fixed

a

100

0

a x

step-fixed

exemplary see example

   100 for x < 10    250 for   10 ≤ x < 20    500 for x ≥ 20

zero in every interval; not differentiable at the jump discontinuities

b

Unit costs c(x)

b

the following applies to the example:  100    x for x < 10   250 x for   10 ≤ x < 20    500 x for x ≥ 20

Table 13.1: Alternative Cost Trends with a ∈ R > 0; b ∈ R > 0; d∈N>1

13 Economic Functions

524 Example 1:

A company manufactures stuffed toys. The material costs to produce a stuffed toy are $6/QU. The rental costs for the operation of the facility including administration are $300 per month. In a considered period, 80 stuffed toys are made. The total costs with fixed costs of $300 and variable costs of $6/QU are therefore: C(x) = C f +Cv (x) = $300 + $6/QU · 80 QU = $780 Example 2: The costs incurred every day of a manufacturing company are as listed: $1, 462.50 at 375 QU and $2, 400 at 1, 000 QU. In order to identify the total cost function, it is necessary to first determine how high the variable costs are. The variable costs represent the non-constant part of the total costs and can therefore be measured between the two different production levels by the difference in production costs: $2, 400 - $1, 462.50 = $937.50 A change in quantity of 1, 000 QU − 375 QU = 625 QU results in an increase in costs of $937.50. The (average) variable unit costs, which correspond to the variable unit costs (= marginal costs) in a linear cost function, are obtained by assigning the change in cost to the change in quantity (linear). $937.50 = $1.50/QU 625 QU If the variable unit costs are multiplied by the quantity produced in one of the two scenarios, the variable costs of this output quantity are calculated as follows: $1.50/QU · 375 QU = $562.50

13.9 Neoclassical Cost Function

525

From a total $1, 462.50 of the production costs, $562.50 comprises the variable (total) costs. Therefore, the fixed costs are: $1, 462.50 − $562.50 = $900. In other words, the total cost function is: C(x) = 900 + 1.5x As a test, the alternative production level should again be analyzed in the same way: $1.50 /QU · 1, 000 QU = $1, 500

From a total $2, 400, $1, 500 comprises the variable (total) costs. The fixed costs are therefore: $2, 400 − $1, 500 = $900.

13.9 Neoclassical Cost Function A neoclassical cost function is a cost function of 2nd degree and is characterized by a disproportionate growth of the total costs C(x) while the quantity produced x increases. The function is convex and strictly monotonically increasing (Fig. 13.14). A distinction is also made between fixed and variable costs in the neoclassical cost function. The variable costs Cv (x) increase disproportionately (progressive; convex) with an increasing quantity x. The fixed costs C f remain unchanged (constant) at any production level.

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13 Economic Functions

Fig. 13.14: Behaviour of the Total Costs Based on a Neoclassical Cost Function dC = C′ (x) which can be calculated based dx on the first derivative of the total cost function C(x) forming a straight line from the origin. This straight line is proportionally increasing. The d 2C second derivative of the neoclassical cost function 2 = C′′ (x) also dx forms a straight line, but parallel to the x-axis (Fig. 13.15). The marginal cost function

13.9 Neoclassical Cost Function

527

Fig. 13.15: The First and Second Derivatives of a Neoclassical Cost Function

Cf are reduced with every unit that is additionally x produced (fixed cost degression). Meanwhile, the variable unit costs Cv (x) increase strictly monotonously (Fig. 13.16). x The fixed unit costs

528

13 Economic Functions

Fig. 13.16: Behaviour of the Variable and Fixed Unit Costs Based on a Neoclassical Cost Function C(x) is at first degressively decreasing and then prox gressively increasing once the (relative) minimum is reached. Since the increasing rate of the variable unit costs is stronger than the decreasing C(x) rate of the fixed unit costs, the total unit cost function is increasing x after the (relative) minimum, as graphically shown in Fig. 13.17. The total unit cost

Fig. 13.17: Behaviour of the Total Unit Costs Based on a Neoclassical Cost Function

13.9 Neoclassical Cost Function

529

The operational optimum describes the output quantity x, where the avC(x) erage total costs (unit costs) are minimal. The corresponding unit x price $/QU determines the long-term lower price limit since this price, according to the full-cost accounting principle, should not fall below this limit (in the long term) in order to avoid losses. If the losses are permanent, (private-sector) production is not a viable option. If the longterm lower price limit is undercut, a direct-cost accounting must be performed to examine whether the price that can be realized in the market is still above the short-term lower price limit and whether it still ensures a positive contribution margin. With a unit price equivalent to the operational optimum, the production of this good generates no profit/loss in the long term. It can be meaningful for a manufacturer to choose the selling price that corresponds to the price at the operational optimum if the manufacturer faces cut-throat competition with this product or if he wants to produce or distribute this product with no intention of making a profit. The operational optimum can be calculated by setting the first derivaC(x) dc(x) tive of the unit cost function, with c(x) = , to zero. If the xdx x value calculated with this method is inserted into the unit cost function, the minimum unit cost can be determined and hence also the longterm lower price limit. Alternatively, the same result can be achieved by dC(x) defining the intersection of the marginal cost curve = C′ (x) with dx C(x) the unit cost curve c(x) = by equating these two functions to each x other. The developments of the total costs, unit costs and marginal costs of a neoclassical cost function are graphically shown in Fig. 13.18.

530

13 Economic Functions

Fig. 13.18: Total Costs, Unit Costs and Marginal Costs of a Neoclassical Cost Function

Example 1: A jewelry company produces necklaces with a manufacturing process that can be represented by the total cost function C(x) = 10 + 0.5x2 . Due to an increase in demand for this product, the company decides to produce more to satisfy the demand. Before the increase in demand, a total cost of $1, 810 per month was generated at a quantity produced of 60 QU per month. However, the demand has now increased by 20 %. How much does the current production cost? 60 QU · 1.2 = 72 QU C(72) = 10 + 0.5 · 722 = $2, 602 With a demand increase of 20 %, the total production costs are now $2, 602, i.e. they have increased by $792 (disproportionately).

13.9 Neoclassical Cost Function

531

Example 2: A watch manufacturer thinks that the average costs he has to pay monthly seem to be relatively high. The (total) cost function for the production of his watches is C(x) = 4, 200 + x2 . To get a more precise overview of the average costs for each watch produced (unit costs), he divides the (total) costs by 60 QU, the average amount of watches that are produced monthly. c(60) =

C(60) 4, 200 + 602 = = $130/QU 60 60

The unit costs of 60 manufactured watches per month are therefore $130/QU. If, however, he wants to produce according to the operational optimum, the minimum of the unit cost function must be identified:

c(x) =

C(x) 4, 200 + x2 4, 200 = = +x x x x dc(x) = −4, 200 · x−2 + 1 dx −4, 200 · x−2 + 1 = 0 x2 = 4, 200 x=

√

4, 200 = 64.81 QU

c(64.81) = $129.61/QU

13 Economic Functions

532

Test: At the operational optimum, the function value at the minimum of the average total costs corresponds to the function value of the corresponding marginal costs: dC(x) = 2·x dx

dC(x) (64.8074) = 2 · 64.8074 = $129.61/QU dx If the watch manufacturer had wanted to optimise the (average) unit costs for this month, he should have produced 64.8074, i.e. 65 watches, instead of 60. His average cost saving per watch would then have been $0.39/QU (= $130/QU − $129.61/QU). Example 3: The finance department of ProductionX Ltd. wants to find out when the average total costs in the company are minimised. Based on a given cost function of C(x) = 200 + 0.2x2 , the operational optimum can be determined as shown below:

c(x) =

C(x) 200 + 0.2x2 200 = = + 0.2 · x x x x dc(x) = −200 · x−2 + 0.2 dx −200 · x−2 + 0.2 = 0 x2 = x=

√

200 = 1, 000 0.2

1, 000 = 31.62 QU

c(31.62) = $12.65/QU Test: At the operational optimum, the function value at the minimum of the average total costs corresponds to the function value of the corre-

13.10 Cost Function According to the Law of Diminishing Returns 533 sponding marginal costs: dC(x) = 0.4 · x dx dC(x) (31.62) = 0.4 · 31.62 = $12.65/QU dx With an output quantity of 31.62 QU, the average total costs of the production discussed here are minimised.

13.10 Cost Function According to the Law of Diminishing Returns A (total) cost function according to the law of diminishing returns represents a cubic function. It begins at the y-intercept, which is determined by the fixed costs, and increases degressively from the y-intercept onwards. It then increases progressively from the (concave/convex) inflection point onwards. Typically, this development of the total cost curve that models diminishing returns can be explained by decreasing growth rates (degressive) at first and increasing growth rates (progressive) of the total costs after the inflection point. The spot at which the inflection point is located is also called the point of diminishing returns. The graph of the (total) cost function in accordance with the law of diminishing returns is first concave and then, from the inflection point, changes to convex. The (total) cost function according to the law of diminishing returns only assumes positive values and does not contain any local (relative) extremes. The curve of this function with the general form C(x) = ax3 + + bx2 + cx + d can be graphically represented as graphically shown in Fig. 13.19.1

1

If there are local extremes in a cubic total cost function, there is a relative maximum on the left of the inflection point and a local minimum on the right. However, the total cost function based on the law of diminishing returns actually has no local extremes by definition. It is strictly monotonically increasing, as long as the inflection point is not a saddle point.

534

13 Economic Functions

Fig. 13.19: Behaviour of the Total Costs Based on a Cost Function According to the Law of Diminishing Returns

A cost function based on the law of diminishing returns with C(x) = ax3 + bx2 + cx + d consists of a variable part Cv (x) = ax3 + bx2 + cx (function of the variable costs) and a fixed part C f = d (fixed costs). The variable costs in this case also first behave degressively with increasing quantity x and then progressively starting from the inflection point of Cv (x). The variable cost function Cv (x) starts at the origin. The fixed costs C f are constant and form a parallel line to the x-axis and has a value of d (Fig. 13.20).

13.10 Cost Function According to the Law of Diminishing Returns 535

Fig. 13.20: Behaviour of the Variable and Fixed Costs Based on a Cost Function According to the Law of Diminishing Return

dC(x) = C′ (x) that associates with the (todx tal) cost function based on the law of diminishing returns also has only positive values. In contrast to the total costs, the marginal cost function has a local minimum: The marginal cost function

dC(x) = 3ax2 + 2bx + c dx

(marginal cost function)

with C(x) = ax3 + bx2 + cx + d

(total cost function)

The minimum of the marginal cost function: 3ax2 + 2bx + c = 0

13 Economic Functions

536

First, the formula must be converted to the normal form so that, for example, by means of the p/q formula the zeros can be determined: x2 + p x1 = − + 2

c 2b x+ =0 3a 3a

r  p 2 2

x1 = −

2b 3a

2

− q with x2 + px + q = 0

v u u +t

b x1 = − + 3a  b A zero is at P1 − + 3a

s

p x2 = − − 2

b 3a

2

r  p 2 2

x2 = −

2b 3a

2

!2

2b 3a

2

s

b 3a

−

c 3a

−

c 3a

2

 c  − 0 . 3a

− q with x2 + px + q = 0

v u u −t

b x2 = − − 3a  b The second zero is at P2 − − 3a

2

s

s

!2

2b 3a

b 3a

b 3a

−

c 3a

−

c 3a

2

2

 c  − 0 . 3a

13.10 Cost Function According to the Law of Diminishing Returns 537 The behaviour of the marginal cost function C′ (x) changes from degressively decreasing before the minimum to progressively increasing after the minimum (Fig. 13.21). It corresponds to a parabola that opens upwards. The total cost function C(x) has a inflection point at the position where the marginal cost function C′ (x) has a minimum.

Fig. 13.21: Behaviour of the Marginal Costs Based on a Cost Function According to the Law of Diminishing Return

C(x) d = ax2 + bx + c + , the cost trend x x is degressively decreasing until it reaches the (relative) minimum and increases then progressively (Fig. 13.22). In the case of the total unit costs

13 Economic Functions

538

Fig. 13.22: Behaviour of the Unit Costs Based on a Cost Function According to the Law of Diminishing Return

At the operational optimum, the average total costs (unit costs) C(x) = c(x) are minimal: x
ax3 + bx2 + cx + d C(x) = c(x) = x x Necessary condition: Sufficient condition:

dc(x) = c′ (x) = 0 dx

d 2 c(x) = c′′ (x) > 0 dx2

If the x-value, at which the unit costs are minimal, is inserted into the unit cost function c(x), the value of the minimum unit costs (y-value) determines the long-term lower price limit in $/QU. This lower price limit should not be undercut in the long term, otherwise losses would be sustained. If a company is at the operational optimum, the entire unit costs including fixed costs are covered. At the operational optimum, the marginal costs of the total cost function are equal to the unit costs:

13.10 Cost Function According to the Law of Diminishing Returns 539 dC(x) = c(x). The corresponding x-value determines the operational dx optimum. The operational minimum describes the output quantity x, where the Cv (x) variable unit costs = cv (x) are minimal: x
ax3 + bx2 + cx cv (x) = x Necessary condition: Sufficient condition:

dcv (x) = cv ′ (x) = 0 dx

d 2 cv (x) = cv ′′ (x) > 0 dx2

The corresponding value of the variable unit costs (y-value) determines the short-term or absolute lower price limit (Fig. 13.23). At the operational minimum, only the total variable unit costs are covered, not the fixed costs. The x-value of the operational minimum can be determined alternatively by equating the marginal costs of the total cost function dC(x) with the variable unit costs: = cv (x). dx

13 Economic Functions

540

Fig. 13.23: Presentation of the Short-term and Long-term Price Lower Limit

Example 1: A production company has a cost function in accordance with the law of diminishing returns C(x) = 0.3x3 − 2x2 + 7x + 16. The head of department wants to find out when exactly the point of diminishing returns occurs, i.e. the point at which the graph changes from degressively increasing to progressively increasing. To determine this, the total cost function must be derived three times: C(x) = 0.3x3 − 2x2 + 7x + 16 C′ (x) = 0.9x2 − 4x + 7 C′′ (x) = 1.8x − 4 C′′′ (x) = 1.8

13.10 Cost Function According to the Law of Diminishing Returns 541 To identify the inflection point (point of diminishing returns), the second derivative must be set to zero (necessary condition):

C′′ (x) = 0 1.8x − 4 = 0 1.8x = 4 x=

20 QU 9

This x-value must now be inserted into the 3rd derivative as a check whether the sufficient condition that the 3rd derivative is not equal to zero is fulfilled:   20 ′′′ C = 1.8 > 0 9 20 QU is equal to zero, it is a saddle point. This 9 is not the case here because: If the 1st derivative x =

C′



20 9



 = 0.9 ·

20 9

2

 −4·

 20 23 +7 = 9 9

The value of the 3rd derivative at the examined position is greater than 20 0, i.e. at the position x = QU, there is a concave/convex inflection 9 point (point of diminishing returns).

13 Economic Functions

542 Example 2:

A production company wants to find out when the short-term lower price limit (operational minimum) and the long-term lower price limit (operational optimum) are reached. The production can be represented by the cost function in terms of the law of diminishing returns: C(x) = x3 − 6x2 + 60x + 100. To determine the operational optimum, the unit cost function c(x) is required:
x3 − 6x2 + 60x + 100 C(x) 100 c(x) = = = x2 − 6x + 60 + x x x The x-value at the operational optimum is then determined by the minimum of the total unit cost function: necessary condition:

c′ (x) = 0

2x − 6 −

100 =0 x2

The left term of this equation can be converted to a standard form by extending the equation on both sides with x2 : 2x3 − 6x2 − 100 = 0 Since this is a cubic function, it requires e.g. a polynomial division. The first zero is x1 = 5.
2x3 − 6x2 − 100 : (x − 5) = 2x2 + 4x + 20

.

13.10 Cost Function According to the Law of Diminishing Returns 543 Afterwards, for instance, the p/q formula can be implemented: p x2 = − + 2

r  p 2 −q 2

2x2 + 4x + 20 = 0 x2 + 2x + 10 = 0 s  2 2 2 x2 = − + − 10 2 2 q x2 = −1 + (1)2 − 10 There are no solutions for x2 and x3 because there is a negative value in the radicand in both cases. The sufficient condition for a minimum of the unit cost function at x1 = 5 d 2 c(x) is = c′′ (x) > 0 dx2 200 c′′ (x) = 2 + 3 x 200 c′′ (5) = 2 + 3 = 3.6 > 0 5 The operational optimum is found at x = 5.

13 Economic Functions

544

Alternatively, the operational optimum can also be identified by equating the marginal costs C′ (x) to the unit costs c(x): C′ (x) = c(x) 3x3 − 12x + 60 = x2 − 6x + 60 + 2x2 − 6x =

100 x

100 x

2x3 − 6x2 − 100 = 0 The calculation procedure is the same as above. There is a zero at x = 5 QU. To determine the operational minimum, the variable unit cost function cv (x) is required:

cv (x) =

Cv (x) x3 − 6x2 + 60x = = x2 − 6x + 60 x x

The x-value of the operational minimum is determined by the minimum variable unit costs: Necessary condition: cv ′ (x) = 0 ⇒ 2x − 6 = 0 ⇔ 2x = 6 x = 3 QU Sufficient condition: cv ′′ (x) > 0 cv ′′ (x) = 2 > 0 The necessary condition is thus fulfilled and the operational minimum corresponds to the quantity produced x = 3 QU.

13.10 Cost Function According to the Law of Diminishing Returns 545 Similarly, the operational minimum can also be identified by equating the marginal costs of the total cost function with the variable unit costs.

C′ (x) = cv (x) 3x2 − 12x + 60 = x2 − 6x + 60 2x2 − 6x = 0 x2 − 3x = 0 p x1 = − + 2

r  p 2 −q 2

 s  −3 −3 2 x1 = − + −0 2 2 

√ x1 = 1.5 + 2.25 − 0 x1 = 1.5 + 1.5 x1 = 3 QU p x2 = − − 2

r  p 2 −q 2

 s  −3 −3 2 x2 = − − −0 2 2 

√ x2 = 1.5 − 2.25 − 0 x2 = 1.5 − 1.5 x2 = 0 QU

13 Economic Functions

546

The operational minimum describes the output quantity x where the variCv (x) able unit costs = cv (x) are minimal: x cv (x) = x2 − 6x + 60 cv (3) = 32 − 18 + 60 = $51/QU cv (0) = 02 − 0 + 60 = $60/QU The operational minimum is at 3 QU.

13.11 Direct Costs versus Indirect Costs In cost accounting, a distinction is made not only between fixed and variable costs, but also between direct costs and indirect costs. Direct costs can be: • Direct manufacturing costs: non material-related manufacturing costs that arise directly during the production process and can be directly allocated to the manufactured product (cost object). Examples of the direct manufacturing costs are labour costs that can be directly assigned to the cost object (direct labour costs per unit, piecework wages), as well as machine costs (machine unit costs) or construction costs. • Direct material costs: material costs that are included in the product that is to be manufactured and can be directly allocated to the product that is manufactured (cost object). Examples of direct material costs are raw materials, auxiliary materials, purchased parts or (preliminary) products. • Special direct costs: special direct costs that are related to the production and distribution. They cannot be directly allocated to a product or (sales-related) service, but to a specific contract or project, for instance. Special direct manufacturing costs are, for example, costs

13.11 Direct Costs versus Indirect Costs

547

for licenses or for special tools required for this (particular) production. Special direct distribution costs can be, for example, (special) packaging costs or commission costs.

Direct costs are characterized by their direct and clear allocation to a cost object, e.g. a product or service, or to a cost centre, e.g. a department, a plant or a company. Accordingly, direct costs are differentiated between direct costs of cost object or direct costs of cost centre. In contrast to direct costs, indirect costs cannot be directly assigned to the reference object. Indirect costs are incurred for multiple end products or orders. They are also called overhead costs. Similar to direct costs, indirect costs can be divided into indirect costs of cost object and indirect costs of cost centre. The following differences in definition are also significant for indirect costs: • False indirect costs / false overheads: indirect costs that could theoretically be allocated directly as direct costs to the cost objects or cost centres, but due to reasons of economic efficiency, are assigned proportionately using classification keys (classification based on indirect costs), e.g. incidentals, electricity costs, lubricant costs. • Primary indirect costs / primary overheads: indirect costs that are generated in the cost centres or cost centre areas themselves due to external acquisitions, i.e. resources or services (e.g. material, personnel, external services) are purchased on the external market. Since the corresponding market prices are known, the primary indirect costs can be clearly identified; there is no problem with value assessment. Consumption documentation, purchase invoices or account statements allow the primary indirect costs to be measured accurately and assigned directly to the cost centres or responsible cost centre areas. • Secondary indirect costs / secondary overheads: While the primary indirect costs are incurred by resources or services outside the company, secondary indirect costs arise from internal service relationships (for example, services provided by the company health insur-

548

13 Economic Functions ance funds). Secondary indirect costs are first determined during internal cost allocation; only then can they be assigned to cost centres or cost centre areas in monetary form. Usually, market prices do not apply to internal activities, so therefore, internal transfer prices must be calculated.

• Indirect manufacturing costs / manufacturing overheads: that part of production costs which cannot be directly allocated. The production costs are composed of direct manufacturing costs (e.g. direct labour costs, direct material costs), indirect manufacturing costs (e.g. salaries for supervisor and technical staff, auxiliary labour costs, costs of auxiliary materials and supplies used in production, electricity costs, imputed depreciation or imputed interest) and special direct manufacturing costs (e.g. special tools, construction plans, patents, licenses). Normally, indirect manufacturing costs cannot be allocated to a single unit produced, but rather to an order or a lot. In the full-cost accounting, the direct manufacturing costs usually form the reference value for the allocation of the indirect manufacturing costs (cost-plus pricing; cost-plus indirect manufacturing costs). • Indirect material costs / material overheads: that part of the material costs that cannot be directly allocated to specific cost objects (products) in the production (e.g. procurement costs, appraisal costs, collective warehousing costs, personnel costs for employees in warehouses or in the purchasing process, depreciation for collective warehouses, i.e. warehouses where other materials are also stored). Indirect material costs are taken into account in (annual) cost center planning. In full-cost accounting, the direct material costs usually form the reference value for the allocation of the indirect material costs (cost-plus pricing; cost-plus indirect material costs). • Indirect administrative costs / administrative overheads: that part of the administrative costs which is incurred in the administration of a company but cannot be allocated to specific products (cost objects) (e.g. salaries for the board of managers, salaries of administrative staff, office supplies, depreciation on business equipment). The indirect administrative costs can be determined within the framework of cost centre accounting as part of the cost distribution sheet. In the unit-based cost unit accounting, the manufacturing costs per

13.11 Direct Costs versus Indirect Costs

549

unit produced forms the reference value for the allocation of the indirect material costs; in the time-based cost unit accounting, the manufacturing costs per unit sold, i.e. the revenue during the observed period (usually annually) forms the reference value for the allocation of the indirect material costs (cost-plus pricing; cost-plus indirect administrative costs). There are various cost allocation principles for assigning costs to the corresponding reference values. A distinction is made between one-dimensional and multi-dimensional cost allocation principles:

13.11.1 One-Dimensional Cost Allocation Principles The (only) reference value here is the activity, i.e. in the case of a (total) cost function C(x) it is the quantity x which is produced or performed. Principle of Causation Only the costs that are additionally incurred in the process of producing this (one) unit can be allocated to the cost object. The costs assigned to a (single) unit of the cost object therefore correspond to their marginal costs. According to the causation principle, the direct costs as well as the variable activity-based indirect costs are allocated to the cost object. An allocation of the fixed activity-based indirect costs is not possible. The causation principle is applied in a German costing methodology, Grenzplankostenrechung, translated as either marginal planned cost accounting or flexible analytic cost planning and accounting.

Principle of Utilisation The costs that are additionally incurred in the process of producing this (one) unit can be allocated to the cost object. The costs assigned to a unit of the cost object thus correspond to their marginal costs and their used capacity costs. The used capacity costs comprise the part of the fixed costs that is allocated to the capacity under usage. The unused part of the fixed costs is called idle capacity costs. The utilisation principle is applied in the activity-based accounting (ABC).

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13 Economic Functions

Principle of Averages In addition to direct costs, the average indirect costs are allocated to the cost object. The average indirect costs are calculated by distributing the total indirect costs linearly by the means of division to the units x which are produced. If the costs cannot be allocated to the cost objects according to the principle of causation in full-cost accounting, they can be distributed according to the principle of averages. The variable costs, which are activity-independent costs, are divided by the number of units produced x and distributed linearly to the units produced. In the case of a single-product company, the total costs C(x) are divided by the total number of units produced x. However, in the case of a multi-product company, the distribution of the indirect costs to the cost objects must be carried out by using a classification based on indirect costs. Principle of Plausibility Those costs that are (plausibly) associated with a (different) cost type are assigned to the cost object. For example, the allocation of the, e.g., indirect material costs, can be oriented to the direct material costs (linear) if there seems to be a plausible (linear) relationship between the direct material costs and the associated indirect material costs. Usually, such an allocation is linear, but a nonlinear relationship between direct costs and indirect costs can also be plausible. Principle of Financial Viability The (pro rata) costs are allocated to the cost object according to its (pro rata) sales revenue. High-revenue (low-revenue) products should be charged with a higher (lower) proportion of the indirect costs or of the total costs (direct and indirect costs). Instead of the sales revenue, either the price or the contribution margin can be chosen for the (proportional) allocation of indirect or total costs.

13.11 Direct Costs versus Indirect Costs

551

13.11.2 Multi-Dimensional Cost Allocation Principles Principle of Decision According to the principle of decision, costs and revenues can only be clearly allocated to each other or to another reference object if they are related to the same (business or operational) decision as the reference object itself. According to this principle, costs are only allocated to a product or service if they are directly caused by the decision to manufacture this product or provide this service.2 Principle of Identity The principle of identity is a further development of the principle of decision. It forms the cost-accounting basis for the relative calculation of direct costs. According to the principle of identity, a business decision is based on three dimensions:3 • Performance | What is to be produced or performed in what quantities? • Organisation | Who is supposed to produce the product(s) or perform the service(s)? • Time | When or to what point in time is this to be produced or performed?

In contrast to the one-dimensional cost allocation principles, the reference objects here are defined in three dimensions, e.g. in the production of goods in the dimensions: • Operational performance (product unit, product type, product group, product line),

2

Cf. Riebel, P. (2013): Einzelkosten- und Deckungsbeitragsrechnung. Grundfragen einer markt- und entscheidungsorientierten Unternehmensrechnung, 6th Edition, Wiesbaden; Schweitzer, M.; Küpper, H.-U.; Friedl, G.; Hofmann, C.; Pedell, B. (2015): Systeme der Kosten- und Erlösrechnung, 11th Edition, Munich. 3 Cf. ibid.

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13 Economic Functions

• Organisational area (company, plant, group, department), • Time period (month, quarter, year). Within these dimensions, the hierarchies of reference objects can possibly be observed through relationships of superordination and subordination. Therefore, the direct costs at a higher decision-making level can also represent indirect costs at a lower decision-making level. Example 1: An employee of ToyFactoryX Ltd. manages to produce a customised toy car in 20 minutes. The costs for the required material are $3/QU and the machine costs are $5/QU. The gross hourly wage for this employee is $24/h. The direct manufacturing costs which are directly allocable to the production of a toy car can be calculated as shown below: 20min/QU min · $24/60 min = $8/QU The direct manufacturing costs are therefore $8/QU + $5/QU = $13/QU. The direct material costs are $3/QU.

Example 2: The rent to be paid per day for the production hall of ToyFactoryX Ltd. is $60/day. These costs cannot be directly allocated to every toy car manufactured and therefore represent indirect costs. The direct costs of $16/QU can be directly assigned to the model produced here; direct manufacturing costs of $13/QU and direct material costs of $3/QU. 50 units of this toy car are made every day. According to the cost-plus pricing, the indirect costs, in this case the rental costs, should be distributed proportionally to the direct costs of this model produced here. The corresponding cost-plus ratio is calculated as shown below:

13.11 Direct Costs versus Indirect Costs

553

$60/day: rent for the production hall (indirect costs per day) $16/QU: direct costs (direct manufacturing costs and direct material costs) per toy car x 50 QU = $300/day (direct costs per day) indirect costs · 100% direct costs $60 = · 100% = 20% $300

Surcharge of total costs =

The total manufacturing costs of these toy cars can thus be calculated ceteris paribus, i.e. without taking other costs into account. The calculation is $16/QU · 1.2 = $19.20/QU.

Example 3: The company ToyFactoryX Ltd. would now like to find out how high the total manufacturing costs are a) per toy car produced and b) per day. The direct manufacturing costs of a toy car are $13/QU. The company is known for the special spray paint for toy cars. This paint is the unique selling proposition (USP) of the company. The cost of painting a toy car is $3/QU. In addition, the total indirect manufacturing costs account for 30 % of the direct manufacturing costs. On a working day, 50 toy cars are made. a) Manufacturing costs for each toy car that is produced: Direct manufacturing costs $13/QU + Special direct manufacturing costs 3/QU + Indirect manufacturing costs $13/QU · 0.3 = $3.9/QU = Manufacturing costs $19.9/QU The manufacturing costs for each toy car that is produced are $19.9/QU.

13 Economic Functions

554

b) Manufacturing costs per day: 50 toy cars are produced daily. $19.9/QU · 50 QU = $995 At ToyFactoryX Ltd., the manufacturing costs are $995 per working day.

13.12 Profit Function A profit function P(x) represents the profit of a company, which depends on the quantity x which is produced or sold. The profit P(x) can be calculated by subtracting the (total) cost C(x) from the revenues R(x): P(x) = R(x) −C(x) P(x) = [p(x) · x] − [C f +Cv (x)] with p = (selling) price and x = quantity produced or sold. If the revenues R(x) exceed the total costs C(x), profits are generated, P(x) > 0. The company is profitable. However, if the total costs C(x) exceed the revenues R(x), negative profits, i.e. losses, are generated, P(x) < 0. The company is unprofitable. The position, i.e. the x-value, at which a profit is earned for the first time, is called the break-even point (BEP). The break-even point is determined by the first zero of the profit function P(x) and marks the beginning of the profit zone. The profit limit is located at the end of the profit zone. There is another zero of the profit function P(x) here. At the break-even point as well as at the profit limit, the costs and the revenues coincide.

555

13.12 Profit Function The break-even point and the profit limit can be determined by: P(x) = 0

or

C(x) = R(x)

At the profit maximum, the x-value of the profit function is where the highest possible profit is made. The profit maximum can be calculated as shown below: necessary condition: P′ (x) = 0 sufficient condition: P′′ (x) < 0 A distinction must be made as shown below for the profit function: I. The price p is constant: This means that the price is fixed, i.e. there is no causality between the selling price p and the quantity sold x, p = constant, p ̸= p(x). The inverse demand function is parallel to the x-axis and the revenue function is represented by a straight line from the origin (Fig. 13.24). See also the chapter on the revenue function.

Fig. 13.24: Behaviour of Profit Based on a Linear Revenue Function

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13 Economic Functions

II. The price p = p(x) is variable: There is a causality between the selling price, p = p(x), and the quantity sold x = x(p). The inverse demand function p = p(x) = −mx + b corresponds to a strictly monotonically decreasing straight line, of which the y-intercept b is determined by x = 0 (prohibitive price) and which intersects the x-axis at p = 0 (saturation quantity). The profit function is represented by a parabola opening downwards, starting at the origin, i.e. at the point (0|0) and ending at the point (xmax |0) (Fig. 13.25). xmax corresponds to the saturation quantity at a price of pmin = 0. The maxib mum of the profit function is at the point with the coordinates x = (2m)    2   b b b −b2 b2 and R = −m · +b = + (2m) (2m) (2m) 4m 2m See also the chapter on the profit function.

Fig. 13.25: Behaviour of Profit Based on a Quadratic Revenue Function

13.12 Profit Function

557

When a company is in a monopolistic position, the Cournot point - which is named after the French economist Antoine Augustin Cournot (18011877) - describes the price-quantity-combination that maximises a monopolist’s profit. The Cournot point represents the monopolistic pricing. Typically, it is located on the left side of the revenue maximum. A smaller quantity of the good x is sold when the profit is maximised than when the revenue is maximised (Fig. 13.26).

Fig. 13.26: The Cournot Point Example 1: The process of producing a quantity of goods x is described as shown below: - The selling price of a good is constant and is $100//QU. - The total costs are calculated with the cost function in accordance with the law of diminishing returns: C(x) = x3 − 3x2 + 52x + 50 [$].

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558

A manufacturer wants to find out when the firm reaches the break-even point and the highest possible profit. To determine the break-even point, the profit limit and the profit maximum, the first step is to define the profit function: P(x) = R(x) −C(x) P(x) = 100x − (x3 − 3x2 + 52x + 50) P(x) = −x3 + 3x2 + 48x − 50 Calculation of the break-even point and the profit limit: P(x) = 0 −x3 + 3x2 + 48x − 50 = 0 As the profit function is a cubic function, a polynomial division, for example, can reveal the possible zeros. A zero is found at x = 1. (−x3 + 3x2 + 48x − 50) : (x − 1) = −x2 + 2x + 50 To determine other possible zeros, the p/q formula can be used:

x2,3

p =− ± 2

r  p 2 −q 2

the following applies for −x2 + 2x + 50 = 0, which is also x2 − 2x − 50 = 0:  s  −2 −2 2 =− ± + 50 2 2 

x2,3

x2,3 = 1 ±

q (−1)2 + 50

559

13.12 Profit Function √ x2 = 1 + 51 ≈ 8.14 √ x3 = 1 − 51 ≈ −6.14 (not economically relevant)

If the positive zeros of the break-even point are interpreted economically, the break-even point is positioned at x = 1 QU. The profit limit is reached at approximately x = 8.14 QU. Calculation of the profit maximum: necessary condition: P′ (x) = 0 P′ (x) = −3x2 + 6x + 48 = 0 x2 − 2x − 16 = 0  s  −2 −2 2 =− ± + 16 2 2 

x1,2

√ x1 = 1 + 1 + 16; √ x1 = 1 + 17 ≈ 5.12;

√ x2 = 1 − 1 + 16

√ x2 = 1 − 17 ≈ −3.12 (not economically relevant)

sufficient condition: P′′ (x) < 0 P′′ (x) = −6x + 6 = 0 √
−6 · 1 + 17 + 6 ≈ −24.74 < 0 The profit-maximising quantity is approximately 5.12 QU. The profit maximum is, √
P 1 + 17 = P(5.12) = −5.123 + 3 · 5.122 + 48 · 5.12 − 50 ≈ $140.19.

13 Economic Functions

560 Example 2:

The monopolist XProducts wants to find out at what quantities the break-even point, the profit limit, the profit maximum and the Cournot point of their current production are located. The process of producing a quantity of goods x is described as shown: • The selling price of a good is variable and is calculated according to the inverse demand function: p(x) = −2x + 100 $/QU. • The total costs are calculated with the cost function in accordance with the law of diminishing returns: C(x) = x3 − 5x2 + 52x + 50 [$].

To determine the break-even point, the profit limit and the profit maximum, the first step is to define the profit function:

P(x) = R(x) −C(x) P(x) = (−2x + 100) · x − x3 − 5x2 + 52x + 50




P(x) = −x3 + 3x2 + 48x − 50 Since P(x) is equal to the profit function of the first example with constant prices, the calculation of the break-even point, the profit limit and the profit maximum is done exactly as demonstrated in the first example (see above). The break-even point is at x = 1 [QU]. The profit limit√is reached at circa x = 8.14 QU. The profit-maximising quantity is 1 + 17 ≈ 5.12 QU. The maximum profit is: √
P 1 + 17 = P (5.12) = −5.123 + 3 · 5.122 + 48 · 5.12 − 50 ≈ $140.19. Calculation of the Cournot point: The Cournot point describes the price-quantity-combination that maximises a monopolist’s profit. The Cournot point represents the monopolistic pricing.

13.12 Profit Function

561

√ pro f it The profit-maximising quantity is at xmax = 1 + 17 ≈ 5.12 QU. The profit-maximising price is calculated as shown below: √
pro f it pmax = −2 · 1 + 17 + 100 ≈ $89.76/QU √ √
Pmax = P 1 + 17 = 2 173 ≈ $140.19 The Cournot point is therefore found at:   √ √

pro f it pro f it xmax | pmax = 1 + 17 | − 2 1 + 17 + 100 ≈ (5.12 | 89.76). The x-value of the Cournot point is typically on the left side of the xvalue at the revenue maximum: R(x) = (−2x + 100) x = −2x2 + 100x necessary condition: R′ (x) = −4x + 100 = 0 ⇒ x = 25 sufficient condition: R′′ (25) = −4 < 0 The revenue-maximising quantity is at: √ revenue = 25 QU > x pro f it = 1 + 17 ≈ 5.12 QU. xmax max A smaller quantity of the good x is sold when the profit is maximised than when the revenue is maximised. The revenue-maximising price is calculated as follows: prevenue = −2 · 25 + 100 = $50/QU max Rmax = R(25) = −2 · 252 + 100 · 25 = $1, 250

Chapter 14

The Peren Theorem The Mathematical Frame in Which We Live Synopsis Humans consume the natural resources of the Earth faster than the Earth is able to regenerate them. Mankind on the whole lives above its means and often at the expense of future generations. Current economic activity, with the aim of maximising monetary profits and generating quantitative growth and prosperity, cannot be continued. The Peren Theorem demonstrates that the consumption of natural resources within a closed system, as represented by Earth, is only possible if their consumption is able to naturally regenerate. If this balance is disturbed for too long a period, this will then result in the natural death of the planet. With an increasing global population, the per capita consumption of natural resources of all humans living on or from the Earth must be proportionately reduced. The Current Human Lifestyle Cannot be Continued Humans consume the natural resources of the Earth faster than the Earth is able to regenerate them. For many years now, human demand for natural resources has exceeded the Earth’s capacity to regenerate these resources. According to the Global Footprint Network 1 , the Earth Overshoot Day in the year 2017 took place on August 2nd of that year.2 In the year before, it was on August 13th , 2016. Mankind on the whole lives above its means and at the expense of future generations. The entire natural resources consumed after Earth Overshoot Day can no longer be replenished by the Earth in the same year concerned. If such an imbalance remains over the long term, then the natural resources of the Earth will be consumed up to the natural death of the planet. 1

Cf. Global Footprint Network (2017): http://www.footprintnetwork.org/, accessed 9 December 2022. 2 Cf. ibid.

© Springer-Verlag GmbH Germany, part of Springer Nature 2023 F. W. Peren, Math for Business and Economics, https://doi.org/10.1007/978-3-662-66975-4_14

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The Living Planet Report 2012 of the World Wide Fund For Nature (WWF)3 shows that roughly up to the year 2030 humans would require two planets in order to cover their need for natural resources if mankind continues to live as it has largely deemed proper thus far. Natural resources are wasted particularly within and for so-called highly developed economies; however, the definition of this concept is misleading because development that results in a human lifestyle that permits the Earth to bear homo sapiens only for approximately seven months per year with itself being consumed for the remaining five months of the same year – both at the expense of other organisms as well as future generations of its own species – can hardly be regarded as “highly developed.” Current economic activity with the aim of maximising monetary profits and generating quantitative growth and prosperity for the companies and citizens of participating national economies already takes place today at the expense of those who either do not or are only partially able to benefit from this predatory ecological exploitation, or who consciously choose to not take part.

The Peren Theorem In the year 2012, the author developed the Peren Theorem4 in discussion with Wiltrud Terlau, director of the International Centre for Sustainable Development (IZNE)5 , and Reiner Clement, professor of economics at Bonn-Rhein-Sieg University in Sankt Augustin, Germany:

“If the users within a closed system employ its natural resources in such measure that its natural regeneration is exceeded over the long term, then the natural environment of this system will be completely exhausted.” 3

World Wide Fund For Nature - WWF (Ed.) (2017): http://wwf.panda.org/about_our_Earth/all_publications/living_planet_report_timeline/lpr_ 2012/, accessed 9 December 2022. 4 Peren, F.W. (2012): The Peren Theorem, New York, unpublished manuscript; Peren, F.W. (2018): Das Peren-Theorem, in: Gadatsch, A. et al. (ed.): Nachhaltiges Wirtschaften im digitalen Zeitalter, Berlin, p. 419-424. 5 International Centre for Sustainable Development – IZNE (2017): https://www.h-brs.de/en/izne, accessed 9 December 2022.

565

14 The Peren Theorem For a closed system stability is6 :

RT ≤ Rregen where

RT = RH + RO N

and

RH =

∑ rI = rH N

I=1

⇔ rH =

RH ∑ rI = N N

with RT

= consumption of natural resources as a whole

Rregen = regeneration of natural resources as a whole RH

= human consumption of natural resources

RO

= consumption of natural resources not caused by humans

rI

= individual per capita consumption of natural resources by humans

rH

= average per capita consumption of natural resources by humans

N

= number of people who live on Earth or access its natural resources

I

= human individuals who live on Earth or access its natural resources; 1, . . . , N

6

Stability is to be understood here as emancipatory in its meaning, i.e. if within a well-defined, temporal interval the inequation RT ≤ Rregen is temporarily violated, then it is nevertheless valid altogether during this period. The scope and location of such a time period are to be selected in such a way that they contain the respectively current point in time and so that the strategic aim of a stable balance between consumed and regenerated natural resources is achieved not only over the longer term, but also for the benefit of those directly affected within the system taken into consideration.

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14 The Peren Theorem

Options for Securing Human Livelihood In relation to mankind and the closed system of the Earth, this mathematical relationship implies that humans have the following options7 in order to secure their existence on Earth: 1. Other consumers of natural resources on this planet are reduced; a practice that mankind already pursues. The habitats of animals and plants are diminished by humans with the consequence that plants and animals are decimated. 2. Mankind reduces itself until this theorem inverts into a positive balance, i.e. until terrestrial consumption caused by humans lies below the natural regeneration of the Earth over the long term. 3. Substantial numbers of mankind leave the Earth. Accordingly, these humans do not use any or hardly any terrestrial natural resources. 4. Mankind modifies the scope and quality of its consumption of natural resources so that this permits a regeneration of natural resources to the extent required. This would require substantial abandonment of the luxury that is understood by large parts of mankind today as prosperity. Individuals would then be entitled to far less natural resources on average than currently claimed and consumed on the average per capita. 5. The recourse to natural resources, i.e. the use, respectively the consumption, of water, soil, air, natural energies and/or sources of energy, plants and animals are put at a clearly higher price than the currently irrational case of disparity in relation to the true value of natural resources. Individual mobility would require a different quality and a clearly higher price. The consumption of meat would have to be more expensive and thus reduced. Global output chains would largely have to be shifted to local production because transportation would have to be priced in accordance with the demand for natural resources. Travel (over long-distances) would also have to be made substantially more expensive and limited.

7

The following list is by no means exhaustive.

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567

6. Mankind substitutes natural resources in favor of synthetic materials; whereas the ecological requirements for the manufacture, transport, recycling and/or disposal of such plastics would also have to be attributed to human consumption of natural raw materials. 7. A more intensive circular economy, i.e. more efficient recycling of already used natural resources could slow down the process of exhaustion of the natural environment of the Earth. However, if increases in efficiency or technical progress results in rebound effects so that increases in efficiency mean that the consumer uses any savings obtained in order to demand more products or services which again consume (additional) natural resources, then increases in efficiency can also result in a so-called backfire, i.e. to rebound effects of more than one hundred percent.

Individual Prosperity Effects The Peren Theorem mathematizes and emancipates a life cycle matter of course. Like every mathematical statement, this theorem is also logically true and thus indisputable in rational terms. If mankind within its terrestrial existence should be interested in a natural environment so that it secures a required (minimum) measure of quality of life for humans – that is certainly evaluated differently by each individual – then operational implementation of this theorem as soon as possible is imperative. Conversely, this theorem also implies that an increasing global population8 has to be accompanied with a proportionate reduction in the average per capita consumption of natural resources if it is to continue to be true that: RT = RH + RO where RH = ∑ rI = rH N.

8

A generally comprehensible overview on population development can be found, for example, on Wikimedia Foundation Inc. (Ed.) (2020): https://en.wikipedia.org/wiki/Population_growth, accessed 9 December 2022, and the literature cited therein.

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Given p percent increase in the global population and unchanged average per capita consumption, then ceteris paribus, i.e. unchanged consumption of natural resources not caused by humans RO the entire consumption of natural resources caused by mankind RH would likewise  p  exhibit proportionate by a factor of 1 + : 100    p p  RH 1 + = rH N 1 + . 100 100 If human consumption of natural resources is meanwhile to be kept constant even with an increasing global population, then the formal relationship of the Peren Theorem9  p  ! RH = rH N 1 + 100 determines the following average per capita consumption of natural resources rH rH =

p −1 RH  1+ N 100

where human consumption of natural resources as a whole RH would remain unchanged in relation to the original state prior to the respectively considered period of increase in the global population. Concomitant with positive population growth of p percent during a certain period, the average per capita consumption of natural resources rH would have to be proportionately reduced by the factor  p −1 1+ 100 In particular, the inhabitants of wealthy national economies, above all the industrialized countries, whose individual human consumption of natural resources rI is clearly above the average per capita consump9

The aim is that human consumption of natural resources altogether RH remains unchanged despite world population growth. Therefore RH is to be equated with p rH N(1 + 100 ), whereby average human per capita consumption of natural resources rH is p −1 ultimately reduced by the growth factor of the world population (1 + 100 ) within the period under consideration.

14 The Peren Theorem

569

tion worldwide rH could by no means continue to maintain their prosperity and lifestyle. If the global population grows meanwhile with unchanged or even increasing (average) prosperity, as understood and lived today, then the consumption of natural resources would additionally accelerate through an (exponentially) increasing global population with simultaneous shortening of the period of total exhaustion of the natural resources of the Earth.

Appendix A

Financial Mathematical Factors

© Springer-Verlag GmbH Germany, part of Springer Nature 2023 F. W. Peren, Math for Business and Economics, https://doi.org/10.1007/978-3-662-66975-4

571

A Financial Mathematical Factors

572 Accumulation Factors qn = (1 + i)n n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

i 0.03 1.0300 1.0609 1.0927 1.1255 1.1593 1.1941 1.2299 1.2668 1.3048 1.3439 1.3842 1.4258 1.4685 1.5126 1.5580 1.6047 1.6528 1.7024 1.7535 1.8061 1.8603 1.9161 1.9736 2.0328 2.0938 2.1566 2.2213 2.2879 2.3566 2.4273 2.5001 2.5751 2.6523 2.7319 2.8139 2.8983 2.9852 3.0748 3.1670 3.2620

0.0375 1.0375 1.0764 1.1168 1.1587 1.2021 1.2472 1.2939 1.3425 1.3928 1.4450 1.4992 1.5555 1.6138 1.6743 1.7371 1.8022 1.8698 1.9399 2.0127 2.0882 2.1665 2.2477 2.3320 2.4194 2.5102 2.6043 2.7020 2.8033 2.9084 3.0175 3.1306 3.2480 3.3698 3.4962 3.6273 3.7633 3.9045 4.0509 4.2028 4.3604

0.04 1.0400 1.0816 1.1249 1.1699 1.2167 1.2653 1.3159 1.3686 1.4233 1.4802 1.5395 1.6010 1.6651 1.7317 1.8009 1.8730 1.9479 2.0258 2.1068 2.1911 2.2788 2.3699 2.4647 2.5633 2.6658 2.7725 2.8834 2.9987 3.1187 3.2434 3.3731 3.5081 3.6484 3.7943 3.9461 4.1039 4.2681 4.4388 4.6164 4.8010

0.0425 1.0425 1.0868 1.1330 1.1811 1.2313 1.2837 1.3382 1.3951 1.4544 1.5162 1.5807 1.6478 1.7179 1.7909 1.8670 1.9463 2.0291 2.1153 2.2052 2.2989 2.3966 2.4985 2.6047 2.7153 2.8308 2.9511 3.0765 3.2072 3.3435 3.4856 3.6338 3.7882 3.9492 4.1171 4.2920 4.4744 4.6646 4.8628 5.0695 5.2850

0.05 1.0500 1.1025 1.1576 1.2155 1.2763 1.3401 1.4071 1.4775 1.5513 1.6289 1.7103 1.7959 1.8856 1.9799 2.0789 2.1829 2.2920 2.4066 2.527 2.6533 2.7860 2.9253 3.0715 3.2251 3.3864 3.5557 3.7335 3.9201 4.1161 4.3219 4.5380 4.7649 5.0032 5.2533 5.5160 5.7918 6.0814 6.3855 6.7048 7.0400

0.06 1.0600 1.1236 1.1910 1.2625 1.3382 1.4185 1.5036 1.5938 1.6895 1.7908 1.8983 2.0122 2.1329 2.2609 2.3966 2.5404 2.6928 2.8543 3.0256 3.2071 3.3996 3.6035 3.8197 4.0489 4.2919 4.5494 4.8223 5.1117 5.4184 5.7435 6.0881 6.4534 6.8406 7.2510 7.6861 8.1473 8.6361 9.1543 9.7035 10.2857

573

A Financial Mathematical Factors Accumulation Factors qn = (1 + i)n n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

i 0.07 1.0700 1.1449 1.2250 1.3108 1.4026 1.5007 1.6058 1.7182 1.8385 1.9672 2.1049 2.2522 2.4098 2.5785 2.7590 2.9522 3.1588 3.3799 3.6165 3.8697 4.1406 4.4304 4.7405 5.0724 5.4274 5.8074 6.2139 6.6488 7.1143 7.6123 8.1451 8.7153 9.3253 9.9781 10.6766 11.4239 12.2236 13.0793 13.9948 14.9745

0.08 1.0800 1.1664 1.2597 1.3605 1.4693 1.5869 1.7138 1.8509 1.9990 2.1589 2.3316 2.5182 2.7196 2.9372 3.1722 3.4259 3.7000 3.9960 4.3157 4.6610 5.0338 5.4365 5.8715 6.3412 6.8485 7.3964 7.9881 8.6271 9.3173 10.0627 10.8677 11.7371 12.6760 13.6901 14.7853 15.9682 17.2456 18.6253 20.1153 21.7245

0.09 1.0900 1.1881 1.2950 1.4116 1.5386 1.6771 1.8280 1.9926 2.1719 2.3674 2.5804 2.8127 3.0658 3.3417 3.6425 3.9703 4.3276 4.7171 5.1417 5.6044 6.1088 6.6586 7.2579 7.9111 8.6231 9.3992 10.2451 11.1671 12.1722 13.2677 14.4618 15.7633 17.1820 18.7284 20.4140 22.2512 24.2538 26.4367 28.8160 31.4094

0.10 1.1000 1.2100 1.3310 1.4641 1.6105 1.7716 1.9487 2.1436 2.3579 2.5937 2.8531 3.1384 3.4523 3.7975 4.1772 4.5950 5.0545 5.5599 6.1159 6.7275 7.4002 8.1403 8.9543 9.8497 10.8347 11.9182 13.1100 14.4210 15.8631 17.4494 19.1943 21.1138 23.2252 25.5477 28.1024 30.9127 34.0039 37.4043 41.1448 45.2593

0.12 1.1200 1.2544 1.4049 1.5735 1.7623 1.9738 2.2107 2.4760 2.7731 3.1058 3.4785 3.8960 4.3635 4.8871 5.4736 6.1304 6.8660 7.6900 8.6128 9.6463 10.8038 12.1003 13.5523 15.1786 17.0001 19.0401 21.3249 23.8839 26.7499 29.9599 33.5551 37.5817 42.0915 47.1425 52.7996 59.1356 66.2318 74.1797 83.0812 93.0510

0.125 1.1250 1.2656 1.4238 1.6018 1.8020 2.0273 2.2807 2.5658 2.8865 3.2473 3.6532 4.1099 4.6236 5.2016 5.8518 6.5833 7.4062 8.3319 9.3734 10.5451 11.8632 13.3461 15.0144 16.8912 19.0026 21.3779 24.0502 27.0564 30.4385 34.2433 38.5237 43.3392 48.7566 54.8512 61.7075 69.4210 78.0986 87.8609 98.8436 111.1990

A Financial Mathematical Factors

574 Accumulation Factors qn = (1 + i)n n 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

i 0.03 3.3599 3.4607 3.5645 3.6715 3.7816 3.8950 4.0119 4.1323 4.2562 4.3839 4.5154 4.6509 4.7904 4.9341 5.0821 5.2346 5.3917 5.5534 5.7200 5.8916 6.0684 6.2504 6.4379 6.6311 6.8300 7.0349 7.2459 7.4633 7.6872 7.9178 8.1554 8.4000 8.6520 8.9116 9.1789 9.4543 9.7379 10.0301 10.3310 10.6409

0.0375 4.5239 4.6935 4.8695 5.0522 5.2416 5.4382 5.6421 5.8537 6.0732 6.3009 6.5372 6.7824 7.0367 7.3006 7.5744 7.8584 8.1531 8.4588 8.7760 9.1051 9.4466 9.8008 10.1684 10.5497 10.9453 11.3557 11.7816 12.2234 12.6818 13.1573 13.6507 14.1626 14.6937 15.2447 15.8164 16.4095 17.0249 17.6633 18.3257 19.0129

0.04 4.9931 5.1928 5.4005 5.6165 5.8412 6.0748 6.3178 6.5705 6.8333 7.1067 7.3910 7.6866 7.9941 8.3138 8.6464 8.9922 9.3519 9.7260 10.1150 10.5196 10.9404 11.3780 11.8332 12.3065 12.7987 13.3107 13.8431 14.3968 14.9727 15.5716 16.1945 16.8423 17.5160 18.2166 18.9453 19.7031 20.4912 21.3108 22.1633 23.0498

0.0425 5.5096 5.7437 5.9878 6.2423 6.5076 6.7842 7.0725 7.3731 7.6865 8.0131 8.3537 8.7087 9.0789 9.4647 9.8670 10.2863 10.7235 11.1792 11.6543 12.1497 12.6660 13.2043 13.7655 14.3505 14.9604 15.5963 16.2591 16.9501 17.6705 18.4215 19.2044 20.0206 20.8715 21.7585 22.6832 23.6473 24.6523 25.7000 26.7922 27.9309

0.05 7.3920 7.7616 8.1497 8.5572 8.9850 9.4343 9.9060 10.4013 10.9213 11.4674 12.0408 12.6428 13.2749 13.9387 14.6356 15.3674 16.1358 16.9426 17.7897 18.6792 19.6131 20.5938 21.6235 22.7047 23.8399 25.0319 26.2835 27.5977 28.9775 30.4264 31.9477 33.5451 35.2224 36.9835 38.8327 40.7743 42.8130 44.9537 47.2014 49.5614

0.06 10.9029 11.5570 12.2505 12.9855 13.7646 14.5905 15.4659 16.3939 17.3775 18.4202 19.5254 20.6969 21.9387 23.2550 24.6503 26.1293 27.6971 29.3589 31.1205 32.9877 34.9670 37.0650 39.2889 41.6462 44.1450 46.7937 49.6013 52.5774 55.7320 59.0759 62.6205 66.3777 70.3604 74.5820 79.0569 83.8003 88.8284 94.1581 99.8075 105.7960

575

A Financial Mathematical Factors Accumulation Factors qn = (1 + i)n n 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

i 0.07 16.0227 17.1443 18.3444 19.6285 21.0025 22.4726 24.0457 25.7289 27.5299 29.4570 31.5190 33.7253 36.0861 38.6122 41.3150 44.2071 47.3015 50.6127 54.1555 57.9464 62.0027 66.3429 70.9869 75.9559 81.2729 86.9620 93.0493 99.5627 106.5321 113.9894 121.9686 130.5065 139.6419 149.4168 159.8760 171.0673 183.0421 195.8550 209.5648 224.2344

0.08 23.4625 25.3395 27.3666 29.5560 31.9204 34.4741 37.2320 40.2106 43.4274 46.9016 50.6537 54.7060 59.0825 63.8091 68.9139 74.4270 80.3811 86.8116 93.7565 101.2571 109.3576 118.1062 127.5547 137.7591 148.7798 160.6822 173.5368 187.4198 202.4133 218.6064 236.0949 254.9825 275.3811 297.4116 321.2045 346.9009 374.6530 404.6252 436.9952 471.9548

0.09 34.2363 37.3175 40.6761 44.3370 48.3273 52.6767 57.4176 62.5852 68.2179 74.3575 81.0497 88.3442 96.2951 104.9617 114.4083 124.7050 135.9285 148.1620 161.4966 176.0313 191.8741 209.1428 227.9656 248.4825 270.8460 295.2221 321.7921 350.7534 382.3212 416.7301 454.2358 495.1170 539.6775 588.2485 641.1909 698.8981 761.7989 830.3608 905.0933 986.5517

0.10 49.7852 54.7637 60.2401 66.2641 72.8905 80.1795 88.1975 97.0172 106.7190 117.3909 129.1299 142.0429 156.2472 171.8719 189.0591 207.9651 228.7616 251.6377 276.8015 304.4816 334.9298 368.4228 405.2651 445.7916 490.3707 539.4078 593.3486 652.6834 717.9518 789.7470 868.7217 955.5938 1051.1532 1156.2685 1271.8954 1399.0849 1538.9934 1692.8927 1862.1820 2048.4002

0.12 104.2171 116.7231 130.7299 146.4175 163.9876 183.6661 205.7061 230.3908 258.0377 289.0022 323.6825 362.5243 406.0273 454.7505 509.3206 570.4391 638.8918 715.5588 801.4258 897.5969 1005.3086 1125.9456 1261.0591 1412.3862 1581.8725 1771.6972 1984.3009 2222.4170 2489.1070 2787.7998 3122.3358 3497.0161 3916.6580 4386.6570 4913.0558 5502.6225 6162.9372 6902.4897 7730.7885 8658.4831

0.125 125.0989 140.7362 158.3283 178.1193 200.3842 225.4322 253.6113 285.3127 320.9768 361.0989 406.2362 457.0157 514.1427 578.4106 650.7119 732.0509 823.5572 926.5019 1042.3146 1172.6039 1319.1794 1484.0769 1669.5865 1878.2848 2113.0704 2377.2042 2674.3547 3008.6490 3384.7301 3807.8214 4283.7991 4819.2740 5421.6832 6099.3936 6861.8178 7719.5450 8684.4882 9770.0492 10991.3054 12365.2185

A Financial Mathematical Factors

576 Accumulation Factors qn = (1 + i)n n 81 82 83 84 85 90 95 100 105 110

i 0.03 10.9601 11.2889 11.6276 11.9764 12.3357 14.3005 16.5782 19.2186 22.2797 25.8282

0.0375 19.7259 20.4656 21.2331 22.0293 22.8554 27.4745 33.0271 39.7018 47.7260 57.3710

0.04 23.9718 24.9307 25.9279 26.9650 28.0436 34.1193 41.5114 50.5049 61.4470 74.7600

0.0425 29.1180 30.3555 31.6456 32.9905 34.3926 42.3493 52.1466 64.2105 79.0650 97.3570

0.05 52.0395 54.6415 57.3736 60.2422 63.2544 80.7304 103.0350 131.5010 167.8300 214.2000

0.06 112.1438 118.8724 126.0047 133.5650 141.5789 189.4645 253.5463 339.3021 454.0630 607.6380

Accumulation Factors qn = (1 + i)n n 81 82 83 84 85 90 95 100 105 110

0.07 239.9308 256.7260 274.6968 293.9255 314.5003 441.1030 618.6697 867.7163 1217.02 1706.93

0.08 509.7112 550.4881 594.5272 642.0893 693.4565 1018.9151 1497.1205 2199.7613 3232.17 4749.12

i 0.09 0.10 1075.3413 2253.2402 1172.1220 2478.5643 1277.6130 2726.4207 1392.5982 2999.0628 1517.9320 3298.9690 2335.5266 5313.0226 3593.4971 8556.6760 5529.0408 13780.6123 8507.11 22193.8 13089.25 35743.4

0.12 9697.5011 10861.2012 12164.5453 13624.2908 15259.2057 26891.9342 47392.7766 83522.2657 147194.8 259407.5

0.125 13910.8708 15649.7297 17605.9459 19806.6891 22282.5253 40153.8341 72358.5129 130392.3900 234971.3 423425.9

577

A Financial Mathematical Factors Discount Factors q−n = (1 + i)−n n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

i 0.03 0.97087 0.94260 0.91514 0.88849 0.86261 0.83748 0.81309 0.78941 0.76642 0.74409 0.72242 0.70138 0.68095 0.66112 0.64186 0.62317 0.60502 0.58739 0.57029 0.55368 0.53755 0.52189 0.50669 0.49193 0.47761 0.46369 0.45019 0.43708 0.42435 0.41199 0.39999 0.38834 0.37703 0.36604 0.35538 0.34503 0.33498 0.32523 0.31575 0.30656

0.0375 0.96386 0.92902 0.89544 0.86307 0.83188 0.80181 0.77283 0.74490 0.71797 0.69202 0.66701 0.64290 0.61966 0.59726 0.57568 0.55487 0.53481 0.51548 0.49685 0.47889 0.46158 0.44490 0.42882 0.41332 0.39838 0.38398 0.37010 0.35672 0.34383 0.33140 0.31942 0.30788 0.29675 0.28603 0.27569 0.26572 0.25612 0.24686 0.23794 0.22934

0.04 0.96154 0.92456 0.88900 0.85480 0.82193 0.79031 0.75992 0.73069 0.70259 0.67556 0.64958 0.62460 0.60057 0.57748 0.55526 0.53391 0.51337 0.49363 0.47464 0.45639 0.43883 0.42196 0.40573 0.39012 0.37512 0.36069 0.34682 0.33348 0.32065 0.30832 0.29646 0.28506 0.27409 0.26355 0.25342 0.24367 0.23430 0.22529 0.21662 0.20829

0.0425 0.95923 0.92013 0.88262 0.84663 0.81212 0.77901 0.74725 0.71679 0.68757 0.65954 0.63265 0.60686 0.58212 0.55839 0.53562 0.51379 0.49284 0.47275 0.45348 0.43499 0.41726 0.40025 0.38393 0.36828 0.35326 0.33886 0.32505 0.31180 0.29908 0.28689 0.27520 0.26398 0.25322 0.24289 0.23299 0.22349 0.21438 0.20564 0.19726 0.18922

0.05 0.95238 0.90703 0.86384 0.82270 0.78353 0.74622 0.71068 0.67684 0.64461 0.61391 0.58468 0.55684 0.53032 0.50507 0.48102 0.45811 0.43630 0.41552 0.39573 0.37689 0.35894 0.34185 0.32557 0.31007 0.29530 0.28124 0.26785 0.25509 0.24295 0.23138 0.22036 0.20987 0.19987 0.19035 0.18129 0.17266 0.16444 0.15661 0.14915 0.14205

0.06 0.94340 0.89000 0.83962 0.79209 0.74726 0.70496 0.66506 0.62741 0.59190 0.55839 0.52679 0.49697 0.46884 0.44230 0.41727 0.39365 0.37136 0.35034 0.33051 0.31180 0.29416 0.27751 0.26180 0.24698 0.23300 0.21981 0.20737 0.19563 0.18456 0.17411 0.16425 0.15496 0.14619 0.13791 0.13011 0.12274 0.11579 0.10924 0.10306 0.09722

A Financial Mathematical Factors

578 Discount Factors q−n = (1 + i)−n n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

i 0.07 0.93458 0.87344 0.81630 0.76290 0.71299 0.66634 0.62275 0.58201 0.54393 0.50835 0.47509 0.44401 0.41496 0.38782 0.36245 0.33873 0.31657 0.29586 0.27651 0.25842 0.24151 0.22571 0.21095 0.19715 0.18425 0.17220 0.16093 0.15040 0.14056 0.13137 0.12277 0.11474 0.10723 0.10022 0.09366 0.08754 0.08181 0.07646 0.07146 0.06678

0.08 0.92593 0.85734 0.79383 0.73503 0.68058 0.63017 0.58349 0.54027 0.50025 0.46319 0.42888 0.39711 0.36770 0.34046 0.31524 0.29189 0.27027 0.25025 0.23171 0.21455 0.19866 0.18394 0.17032 0.15770 0.14602 0.13520 0.12519 0.11591 0.10733 0.09938 0.09202 0.08520 0.07889 0.07305 0.06763 0.06262 0.05799 0.05369 0.04971 0.04603

0.09 0.91743 0.84168 0.77218 0.70843 0.64993 0.59627 0.54703 0.50187 0.46043 0.42241 0.38753 0.35553 0.32618 0.29925 0.27454 0.25187 0.23107 0.21199 0.19449 0.17843 0.16370 0.15018 0.13778 0.12640 0.11597 0.10639 0.09761 0.08955 0.08215 0.07537 0.06915 0.06344 0.05820 0.05339 0.04899 0.04494 0.04123 0.03783 0.03470 0.03184

0.10 0.90909 0.82645 0.75131 0.68301 0.62092 0.56447 0.51316 0.46651 0.42410 0.38554 0.35049 0.31863 0.28966 0.26333 0.23939 0.21763 0.19784 0.17986 0.16351 0.14864 0.13513 0.12285 0.11168 0.10153 0.09230 0.08391 0.07628 0.06934 0.06304 0.05731 0.05210 0.04736 0.04306 0.03914 0.03558 0.03235 0.02941 0.02673 0.02430 0.02209

0.12 0.89286 0.79719 0.71178 0.63552 0.56743 0.50663 0.45235 0.40388 0.36061 0.32197 0.28748 0.25668 0.22917 0.20462 0.18270 0.16312 0.14564 0.13004 0.11611 0.10367 0.09256 0.08264 0.07379 0.06588 0.05882 0.05252 0.04689 0.04187 0.03738 0.03338 0.02980 0.02661 0.02376 0.02121 0.01894 0.01691 0.01510 0.01348 0.01204 0.01075

0.125 0.88889 0.79012 0.70233 0.62430 0.55493 0.49327 0.43846 0.38974 0.34644 0.30795 0.27373 0.24332 0.21628 0.19225 0.17089 0.15190 0.13502 0.12002 0.10668 0.09483 0.08429 0.07493 0.06660 0.05920 0.05262 0.04678 0.04158 0.03696 0.03285 0.02920 0.02596 0.02307 0.02051 0.01823 0.01621 0.01440 0.01280 0.01138 0.01012 0.00899

579

A Financial Mathematical Factors Discount Factors q−n = (1 + i)−n n 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

i 0.03 0.29763 0.28896 0.28054 0.27237 0.26444 0.25674 0.24926 0.24200 0.23495 0.22811 0.22146 0.21501 0.20875 0.20267 0.19677 0.19104 0.18547 0.18007 0.17483 0.16973 0.16479 0.15999 0.15533 0.15081 0.14641 0.14215 0.13801 0.13399 0.13009 0.12630 0.12262 0.11905 0.11558 0.11221 0.10895 0.10577 0.10269 0.09970 0.09680 0.09398

0.0375 0.22105 0.21306 0.20536 0.19794 0.19078 0.18389 0.17724 0.17083 0.16466 0.15871 0.15297 0.14744 0.14211 0.13698 0.13202 0.12725 0.12265 0.11822 0.11395 0.10983 0.10586 0.10203 0.09834 0.09479 0.09136 0.08806 0.08488 0.08181 0.07885 0.07600 0.07326 0.07061 0.06806 0.06560 0.06323 0.06094 0.05874 0.05661 0.05457 0.05260

0.04 0.20028 0.19257 0.18517 0.17805 0.17120 0.16461 0.15828 0.15219 0.14634 0.14071 0.13530 0.13010 0.12509 0.12028 0.11566 0.11121 0.10693 0.10282 0.09886 0.09506 0.09140 0.08789 0.08451 0.08126 0.07813 0.07513 0.07224 0.06946 0.06679 0.06422 0.06175 0.05937 0.05709 0.05490 0.05278 0.05075 0.04880 0.04692 0.04512 0.04338

0.0425 0.18150 0.17410 0.16700 0.16020 0.15367 0.14740 0.14139 0.13563 0.13010 0.12479 0.11971 0.11483 0.11015 0.10566 0.10135 0.09722 0.09325 0.08945 0.08580 0.08231 0.07895 0.07573 0.07265 0.06968 0.06684 0.06412 0.06150 0.05900 0.05659 0.05428 0.05207 0.04995 0.04791 0.04596 0.04409 0.04229 0.04056 0.03891 0.03732 0.03580

0.05 0.13528 0.12884 0.12270 0.11686 0.11130 0.10600 0.10095 0.09614 0.09156 0.08720 0.08305 0.07910 0.07533 0.07174 0.06833 0.06507 0.06197 0.05902 0.05621 0.05354 0.05099 0.04856 0.04625 0.04404 0.04195 0.03995 0.03805 0.03623 0.03451 0.03287 0.03130 0.02981 0.02839 0.02704 0.02575 0.02453 0.02336 0.02225 0.02119 0.02018

0.06 0.09172 0.08653 0.08163 0.07701 0.07265 0.06854 0.06466 0.06100 0.05755 0.05429 0.05122 0.04832 0.04558 0.04300 0.04057 0.03827 0.03610 0.03406 0.03213 0.03031 0.02860 0.02698 0.02545 0.02401 0.02265 0.02137 0.02016 0.01902 0.01794 0.01693 0.01597 0.01507 0.01421 0.01341 0.01265 0.01193 0.01126 0.01062 0.01002 0.00945

A Financial Mathematical Factors

580 Discount Factors q−n = (1 + i)−n n 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

i 0.07 0.06241 0.05833 0.05451 0.05095 0.04761 0.04450 0.04159 0.03887 0.03632 0.03395 0.03173 0.02965 0.02771 0.02590 0.02420 0.02262 0.02114 0.01976 0.01847 0.01726 0.01613 0.01507 0.01409 0.01317 0.01230 0.01150 0.01075 0.01004 0.00939 0.00877 0.00820 0.00766 0.00716 0.00669 0.00625 0.00585 0.00546 0.00511 0.00477 0.00446

0.08 0.04262 0.03946 0.03654 0.03383 0.03133 0.02901 0.02686 0.02487 0.02303 0.02132 0.01974 0.01828 0.01693 0.01567 0.01451 0.01344 0.01244 0.01152 0.01067 0.00988 0.00914 0.00847 0.00784 0.00726 0.00672 0.00622 0.00576 0.00534 0.00494 0.00457 0.00424 0.00392 0.00363 0.00336 0.00311 0.00288 0.00267 0.00247 0.00229 0.00212

0.09 0.02921 0.02680 0.02458 0.02255 0.02069 0.01898 0.01742 0.01598 0.01466 0.01345 0.01234 0.01132 0.01038 0.00953 0.00874 0.00802 0.00736 0.00675 0.00619 0.00568 0.00521 0.00478 0.00439 0.00402 0.00369 0.00339 0.00311 0.00285 0.00262 0.00240 0.00220 0.00202 0.00185 0.00170 0.00156 0.00143 0.00131 0.00120 0.00110 0.00101

0.10 0.02009 0.01826 0.01660 0.01509 0.01372 0.01247 0.01134 0.01031 0.00937 0.00852 0.00774 0.00704 0.00640 0.00582 0.00529 0.00481 0.00437 0.00397 0.00361 0.00328 0.00299 0.00271 0.00247 0.00224 0.00204 0.00185 0.00169 0.00153 0.00139 0.00127 0.00115 0.00105 0.00095 0.00086 0.00079 0.00071 0.00065 0.00059 0.00054 0.00049

0.12 0.00960 0.00857 0.00765 0.00683 0.00610 0.00544 0.00486 0.00434 0.00388 0.00346 0.00309 0.00276 0.00246 0.00220 0.00196 0.00175 0.00157 0.00140 0.00125 0.00111 0.00099 0.00089 0.00079 0.00071 0.00063 0.00056 0.00050 0.00045 0.00040 0.00036 0.00032 0.00029 0.00026 0.00023 0.00020 0.00018 0.00016 0.00014 0.00013 0.00012

0.125 0.00799 0.00711 0.00632 0.00561 0.00499 0.00444 0.00394 0.00350 0.00312 0.00277 0.00246 0.00219 0.00194 0.00173 0.00154 0.00137 0.00121 0.00108 0.00096 0.00085 0.00076 0.00067 0.00060 0.00053 0.00047 0.00042 0.00037 0.00033 0.00030 0.00026 0.00023 0.00021 0.00018 0.00016 0.00015 0.00013 0.00012 0.00010 0.00009 0.00008

581

A Financial Mathematical Factors Discount Factors q−n = (1 + i)−n n 81 82 83 84 85 90 95 100 105 110

i 0.03 0.09124 0.08858 0.08600 0.08350 0.08107 0.06993 0.06032 0.05203 0.04488 0.03872

0.0375 0.05069 0.04886 0.04710 0.04539 0.04375 0.03640 0.03028 0.02519 0.02095 0.01743

0.04 0.04172 0.04011 0.03857 0.03709 0.03566 0.02931 0.02409 0.01980 0.01627 0.01338

0.0425 0.03434 0.03294 0.03160 0.03031 0.02908 0.02361 0.01918 0.01557 0.01265 0.01027

0.05 0.01922 0.01830 0.01743 0.01660 0.01581 0.01239 0.00971 0.00760 0.00596 0.00467

0.06 0.00892 0.00841 0.00794 0.00749 0.00706 0.00528 0.00394 0.00295 0.00220 0.00165

Discount Factors q−n = (1 + i)−n n 81 82 83 84 85 90 95 100 105 110

i 0.07 0.00417 0.00390 0.00364 0.00340 0.00318 0.00227 0.00162 0.00115 0.00082 0.00059

0.08 0.00196 0.00182 0.00168 0.00156 0.00144 0.00098 0.00067 0.00045 0.00031 0.00021

0.09 0.00093 0.00085 0.00078 0.00072 0.00066 0.00043 0.00028 0.00018 0.00012 0.00008

0.10 0.00044 0.00040 0.00037 0.00033 0.00030 0.00019 0.00012 0.00007 0.00005 0.00003

0.12 0.00010 0.00009 0.00008 0.00007 0.00007 0.00004 0.00002 0.00001 0.00001 0.00000

0.125 0.00007 0.00006 0.00006 0.00005 0.00004 0.00002 0.00001 0.00001 0.00000 0.00000

A Financial Mathematical Factors

582 Repayment Factor n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

q−1 qn −1

=

i (1+i)n −1

i 0.03 1.00000 0.49261 0.32353 0.23903 0.18835 0.15460 0.13051 0.11246 0.09843 0.08723 0.07808 0.07046 0.06403 0.05853 0.05377 0.04961 0.04595 0.04271 0.03981 0.03722 0.03487 0.03275 0.03081 0.02905 0.02743 0.02594 0.02456 0.02329 0.02211 0.02102 0.02000 0.01905 0.01816 0.01732 0.01654 0.01580 0.01511 0.01446 0.01384 0.01326

0.0375 1.00000 0.49080 0.32114 0.23637 0.18555 0.15171 0.12757 0.10950 0.09547 0.08426 0.07512 0.06751 0.06110 0.05561 0.05088 0.04674 0.04311 0.03990 0.03703 0.03446 0.03215 0.03006 0.02815 0.02642 0.02483 0.02337 0.02203 0.02080 0.01965 0.01859 0.01760 0.01668 0.01582 0.01502 0.01427 0.01357 0.01291 0.01229 0.01171 0.01116

0.04 1.00000 0.49020 0.32035 0.23549 0.18463 0.15076 0.12661 0.10853 0.09449 0.08329 0.07415 0.06655 0.06014 0.05467 0.04994 0.04582 0.04220 0.03899 0.03614 0.03358 0.03128 0.02920 0.02731 0.02559 0.02401 0.02257 0.02124 0.02001 0.01888 0.01783 0.01686 0.01595 0.01510 0.01431 0.01358 0.01289 0.01224 0.01163 0.01106 0.01052

0.0425 1.00000 0.48960 0.31956 0.23462 0.18371 0.14982 0.12565 0.10756 0.09353 0.08233 0.07319 0.06560 0.05920 0.05374 0.04902 0.04491 0.04130 0.03811 0.03526 0.03272 0.03043 0.02836 0.02649 0.02478 0.02321 0.02178 0.02047 0.01925 0.01813 0.01710 0.01614 0.01524 0.01441 0.01363 0.01291 0.01223 0.01160 0.01100 0.01044 0.00992

0.05 1.00000 0.48780 0.31721 0.23201 0.18097 0.14702 0.12282 0.10472 0.09069 0.07950 0.07039 0.06283 0.05646 0.05102 0.04634 0.04227 0.03870 0.03555 0.03275 0.03024 0.02800 0.02597 0.02414 0.02247 0.02095 0.01956 0.01829 0.01712 0.01605 0.01505 0.01413 0.01328 0.01249 0.01176 0.01107 0.01043 0.00984 0.00928 0.00876 0.00828

0.06 1.00000 0.48544 0.31411 0.22859 0.17740 0.14336 0.11914 0.10104 0.08702 0.07587 0.06679 0.05928 0.05296 0.04758 0.04296 0.03895 0.03544 0.03236 0.02962 0.02718 0.02500 0.02305 0.02128 0.01968 0.01823 0.01690 0.01570 0.01459 0.01358 0.01265 0.01179 0.01100 0.01027 0.00960 0.00897 0.00839 0.00786 0.00736 0.00689 0.00646

583

A Financial Mathematical Factors Repayment Factor n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

q−1 qn −1

=

i (1+i)n −1

i 0.07 1.00000 0.48309 0.31105 0.22523 0.17389 0.13980 0.11555 0.09747 0.08349 0.07238 0.06336 0.05590 0.04965 0.04434 0.03979 0.03586 0.03243 0.02941 0.02675 0.02439 0.02229 0.02041 0.01871 0.01719 0.01581 0.01456 0.01343 0.01239 0.01145 0.01059 0.00980 0.00907 0.00841 0.00780 0.00723 0.00672 0.00624 0.00580 0.00539 0.00501

0.08 1.00000 0.48077 0.30803 0.22192 0.17046 0.13632 0.11207 0.09401 0.08008 0.06903 0.06008 0.05270 0.04652 0.04130 0.03683 0.03298 0.02963 0.02670 0.02413 0.02185 0.01983 0.01803 0.01642 0.01498 0.01368 0.01251 0.01145 0.01049 0.00962 0.00883 0.00811 0.00745 0.00685 0.00630 0.00580 0.00534 0.00492 0.00454 0.00419 0.00386

0.09 1.00000 0.47847 0.30505 0.21867 0.16709 0.13292 0.10869 0.09067 0.07680 0.06582 0.05695 0.04965 0.04357 0.03843 0.03406 0.03030 0.02705 0.02421 0.02173 0.01955 0.01762 0.01590 0.01438 0.01302 0.01181 0.01072 0.00973 0.00885 0.00806 0.00734 0.00669 0.00610 0.00556 0.00508 0.00464 0.00424 0.00387 0.00354 0.00324 0.00296

0.10 1.00000 0.47619 0.30211 0.21547 0.16380 0.12961 0.10541 0.08744 0.07364 0.06275 0.05396 0.04676 0.04078 0.03575 0.03147 0.02782 0.02466 0.02193 0.01955 0.01746 0.01562 0.01401 0.01257 0.01130 0.01017 0.00916 0.00826 0.00745 0.00673 0.00608 0.00550 0.00497 0.00450 0.00407 0.00369 0.00334 0.00303 0.00275 0.00249 0.00226

0.12 1.00000 0.47170 0.29635 0.20923 0.15741 0.12323 0.09912 0.08130 0.06768 0.05698 0.04842 0.04144 0.03568 0.03087 0.02682 0.02339 0.02046 0.01794 0.01576 0.01388 0.01224 0.01081 0.00956 0.00846 0.00750 0.00665 0.00590 0.00524 0.00466 0.00414 0.00369 0.00328 0.00292 0.00260 0.00232 0.00206 0.00184 0.00164 0.00146 0.00130

0.125 1.00000 0.47059 0.29493 0.20771 0.15585 0.12168 0.09760 0.07983 0.06626 0.05562 0.04711 0.04019 0.03450 0.02975 0.02576 0.02239 0.01951 0.01705 0.01493 0.01310 0.01151 0.01012 0.00892 0.00787 0.00694 0.00613 0.00542 0.00480 0.00425 0.00376 0.00333 0.00295 0.00262 0.00232 0.00206 0.00183 0.00162 0.00144 0.00128 0.00113

A Financial Mathematical Factors

584 Repayment Factor n 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

q−1 qn −1

=

i (1+i)n −1

i 0.03 0.01271 0.01219 0.01170 0.01123 0.01079 0.01036 0.00996 0.00958 0.00921 0.00887 0.00853 0.00822 0.00791 0.00763 0.00735 0.00708 0.00683 0.00659 0.00636 0.00613 0.00592 0.00571 0.00552 0.00533 0.00515 0.00497 0.00480 0.00464 0.00449 0.00434 0.00419 0.00405 0.00392 0.00379 0.00367 0.00355 0.00343 0.00332 0.00322 0.00311

0.0375 0.01064 0.01015 0.00969 0.00925 0.00884 0.00845 0.00808 0.00773 0.00739 0.00707 0.00677 0.00649 0.00621 0.00595 0.00570 0.00547 0.00524 0.00503 0.00482 0.00463 0.00444 0.00426 0.00409 0.00393 0.00377 0.00362 0.00348 0.00334 0.00321 0.00308 0.00296 0.00285 0.00274 0.00263 0.00253 0.00243 0.00234 0.00225 0.00216 0.00208

0.04 0.01002 0.00954 0.00909 0.00866 0.00826 0.00788 0.00752 0.00718 0.00686 0.00655 0.00626 0.00598 0.00572 0.00547 0.00523 0.00500 0.00479 0.00458 0.00439 0.00420 0.00402 0.00385 0.00369 0.00354 0.00339 0.00325 0.00311 0.00299 0.00286 0.00275 0.00263 0.00252 0.00242 0.00232 0.00223 0.00214 0.00205 0.00197 0.00189 0.00181

0.0425 0.00942 0.00896 0.00852 0.00811 0.00772 0.00735 0.00700 0.00667 0.00636 0.00606 0.00578 0.00551 0.00526 0.00502 0.00479 0.00458 0.00437 0.00418 0.00399 0.00381 0.00364 0.00348 0.00333 0.00318 0.00304 0.00291 0.00279 0.00266 0.00255 0.00244 0.00233 0.00223 0.00214 0.00205 0.00196 0.00188 0.00180 0.00172 0.00165 0.00158

0.05 0.00782 0.00739 0.00699 0.00662 0.00626 0.00593 0.00561 0.00532 0.00504 0.00478 0.00453 0.00429 0.00407 0.00386 0.00367 0.00348 0.00330 0.00314 0.00298 0.00283 0.00269 0.00255 0.00242 0.00230 0.00219 0.00208 0.00198 0.00188 0.00179 0.00170 0.00162 0.00154 0.00146 0.00139 0.00132 0.00126 0.00120 0.00114 0.00108 0.00103

0.06 0.00606 0.00568 0.00533 0.00501 0.00470 0.00441 0.00415 0.00390 0.00366 0.00344 0.00324 0.00305 0.00287 0.00270 0.00254 0.00239 0.00225 0.00212 0.00199 0.00188 0.00177 0.00166 0.00157 0.00148 0.00139 0.00131 0.00123 0.00116 0.00110 0.00103 0.00097 0.00092 0.00087 0.00082 0.00077 0.00072 0.00068 0.00064 0.00061 0.00057

585

A Financial Mathematical Factors Repayment Factor n 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

q−1 qn −1

=

i (1+i)n −1

i 0.07 0.00466 0.00434 0.00404 0.00376 0.00350 0.00326 0.00304 0.00283 0.00264 0.00246 0.00229 0.00214 0.00200 0.00186 0.00174 0.00162 0.00151 0.00141 0.00132 0.00123 0.00115 0.00107 0.00100 0.00093 0.00087 0.00081 0.00076 0.00071 0.00066 0.00062 0.00058 0.00054 0.00050 0.00047 0.00044 0.00041 0.00038 0.00036 0.00034 0.00031

0.08 0.00356 0.00329 0.00303 0.00280 0.00259 0.00239 0.00221 0.00204 0.00189 0.00174 0.00161 0.00149 0.00138 0.00127 0.00118 0.00109 0.00101 0.00093 0.00086 0.00080 0.00074 0.00068 0.00063 0.00058 0.00054 0.00050 0.00046 0.00043 0.00040 0.00037 0.00034 0.00031 0.00029 0.00027 0.00025 0.00023 0.00021 0.00020 0.00018 0.00017

0.09 0.00271 0.00248 0.00227 0.00208 0.00190 0.00174 0.00160 0.00146 0.00134 0.00123 0.00112 0.00103 0.00094 0.00087 0.00079 0.00073 0.00067 0.00061 0.00056 0.00051 0.00047 0.00043 0.00040 0.00036 0.00033 0.00031 0.00028 0.00026 0.00024 0.00022 0.00020 0.00018 0.00017 0.00015 0.00014 0.00013 0.00012 0.00011 0.00010 0.00009

0.10 0.00205 0.00186 0.00169 0.00153 0.00139 0.00126 0.00115 0.00104 0.00095 0.00086 0.00078 0.00071 0.00064 0.00059 0.00053 0.00048 0.00044 0.00040 0.00036 0.00033 0.00030 0.00027 0.00025 0.00022 0.00020 0.00019 0.00017 0.00015 0.00014 0.00013 0.00012 0.00010 0.00010 0.00009 0.00008 0.00007 0.00007 0.00006 0.00005 0.00005

0.12 0.00116 0.00104 0.00092 0.00083 0.00074 0.00066 0.00059 0.00052 0.00047 0.00042 0.00037 0.00033 0.00030 0.00026 0.00024 0.00021 0.00019 0.00017 0.00015 0.00013 0.00012 0.00011 0.00010 0.00009 0.00008 0.00007 0.00006 0.00005 0.00005 0.00004 0.00004 0.00003 0.00003 0.00003 0.00002 0.00002 0.00002 0.00002 0.00002 0.00001

0.125 0.00101 0.00089 0.00079 0.00071 0.00063 0.00056 0.00049 0.00044 0.00039 0.00035 0.00031 0.00027 0.00024 0.00022 0.00019 0.00017 0.00015 0.00014 0.00012 0.00011 0.00009 0.00008 0.00007 0.00007 0.00006 0.00005 0.00005 0.00004 0.00004 0.00003 0.00003 0.00003 0.00002 0.00002 0.00002 0.00002 0.00001 0.00001 0.00001 0.00001

A Financial Mathematical Factors

586 Repayment Factor n 85 90 95 100 105

85 90 95 100 105

=

i (1+i)n −1

i 0.03 0.00265 0.00226 0.00193 0.00165 0.00141

0.0375 0.00172 0.00142 0.00117 0.00097 0.00080

Repayment Factor n

q−1 qn −1

q−1 qn −1

0.04 0.00148 0.00121 0.00099 0.00081 0.00066

=

0.0425 0.00127 0.00103 0.00083 0.00067 0.00054

0.05 0.0008 0.00063 0.00049 0.00038 0.00030

0.06 0.00043 0.00032 0.00024 0.00018 0.00013

i (1+i)n −1

i 0.07 0.00022 0.00016 0.00011 0.00008 0.00006

0.08 0.00012 0.00008 0.00005 0.00004 0.00002

0.09 0.00006 0.00004 0.00003 0.00002 0.00001

0.10 0.00003 0.00002 0.00001 0.00001 0.00000

0.12 0.00001 0.00000 0.00000 0.00000 0.00000

0.125 0.00001 0.00000 0.00000 0.00000 0.00000

587

A Financial Mathematical Factors qn −1 qn ·(q−1)

Annuity Value Factors (in advance) n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

·q =

(1+i)n −1 i·(1+i)n

· (1 + i)

i 0.03 1.00000 1.97087 2.91347 3.82861 4.71710 5.57971 6.41719 7.23028 8.01969 8.78611 9.53020 10.25262 10.95400 11.63496 12.29607 12.93794 13.56110 14.16612 14.75351 15.32380 15.87747 16.41502 16.93692 17.44361 17.93554 18.41315 18.87684 19.32703 19.76411 20.18845 20.60044 21.00043 21.38877 21.76579 22.13184 22.48722 22.83225 23.16724 23.49246 23.80822

0.0375 1.00000 1.96386 2.89287 3.78831 4.65138 5.48326 6.28507 7.05790 7.80280 8.52077 9.21279 9.87979 10.52269 11.14236 11.73962 12.31530 12.87017 13.40498 13.92046 14.41731 14.89620 15.35779 15.80269 16.23151 16.64482 17.04320 17.42718 17.79729 18.15401 18.49784 18.82925 19.14867 19.45655 19.75330 20.03933 20.31501 20.58074 20.83685 21.08371 21.32165

0.04 1.00000 1.96154 2.88609 3.77509 4.62990 5.45182 6.24214 7.00205 7.73274 8.43533 9.11090 9.76048 10.38507 10.98565 11.56312 12.11839 12.65230 13.16567 13.65930 14.13394 14.59033 15.02916 15.45112 15.85684 16.24696 16.62208 16.98277 17.32959 17.66306 17.98371 18.29203 18.58849 18.87355 19.14765 19.41120 19.66461 19.90828 20.14258 20.36786 20.58448

0.0425 1.00000 1.95923 2.87936 3.76198 4.60861 5.42073 6.19974 6.94699 7.66378 8.35135 9.01089 9.64354 10.25039 10.83251 11.39090 11.92652 12.44031 12.93315 13.40590 13.85938 14.29437 14.71162 15.11187 15.49580 15.86407 16.21734 16.55620 16.88124 17.19304 17.49213 17.77902 18.05421 18.31819 18.57141 18.81430 19.04729 19.27078 19.48516 19.69080 19.88806

0.05 1.00000 1.95238 2.85941 3.72325 4.54595 5.32948 6.07569 6.78637 7.46321 8.10782 8.72173 9.30641 9.86325 10.39357 10.89864 11.37966 11.83777 12.27407 12.68959 13.08532 13.46221 13.82115 14.16300 14.48857 14.79864 15.09394 15.37519 15.64303 15.89813 16.14107 16.37245 16.59281 16.80268 17.00255 17.19290 17.37419 17.54685 17.71129 17.86789 18.01704

0.06 1.00000 1.94340 2.83339 3.67301 4.46511 5.21236 5.91732 6.58238 7.20979 7.80169 8.36009 8.88687 9.38384 9.85268 10.29498 10.71225 11.10590 11.47726 11.82760 12.15812 12.46992 12.76408 13.04158 13.30338 13.55036 13.78336 14.00317 14.21053 14.40616 14.59072 14.76483 14.92909 15.08404 15.23023 15.36814 15.49825 15.62099 15.73678 15.84602 15.94907

A Financial Mathematical Factors

588 Annuity Value Factors (in advance) n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

qn −1 qn ·(q−1)

·q =

(1+i)n −1 i·(1+i)n

· (1 + i)

i 0.07 1.00000 1.93458 2.80802 3.62432 4.38721 5.10020 5.76654 6.38929 6.97130 7.51523 8.02358 8.49867 8.94269 9.35765 9.74547 10.10791 10.44665 10.76322 11.05909 11.33560 11.59401 11.83553 12.06124 12.27219 12.46933 12.65358 12.82578 12.98671 13.13711 13.27767 13.40904 13.53181 13.64656 13.75379 13.85401 13.94767 14.03521 14.11702 14.19347 14.26493

0.08 1.00000 1.92593 2.78326 3.57710 4.31213 4.99271 5.62288 6.20637 6.74664 7.24689 7.71008 8.13896 8.53608 8.90378 9.24424 9.55948 9.85137 10.12164 10.37189 10.60360 10.81815 11.01680 11.20074 11.37106 11.52876 11.67478 11.80998 11.93516 12.05108 12.15841 12.25778 12.34980 12.43500 12.51389 12.58693 12.65457 12.71719 12.77518 12.82887 12.87858

0.09 1.00000 1.91743 2.75911 3.53129 4.23972 4.88965 5.48592 6.03295 6.53482 6.99525 7.41766 7.80519 8.16073 8.48690 8.78615 9.06069 9.31256 9.54363 9.75563 9.95011 10.12855 10.29224 10.44243 10.58021 10.70661 10.82258 10.92897 11.02658 11.11613 11.19828 11.27365 11.34280 11.40624 11.46444 11.51784 11.56682 11.61176 11.65299 11.69082 11.72552

0.10 1.00000 1.90909 2.73554 3.48685 4.16987 4.79079 5.35526 5.86842 6.33493 6.75902 7.14457 7.49506 7.81369 8.10336 8.36669 8.60608 8.82371 9.02155 9.20141 9.36492 9.51356 9.64869 9.77154 9.88322 9.98474 10.07704 10.16095 10.23722 10.30657 10.36961 10.42691 10.47901 10.52638 10.56943 10.60857 10.64416 10.67651 10.70592 10.73265 10.75696

0.12 1.00000 1.89286 2.69005 3.40183 4.03735 4.60478 5.11141 5.56376 5.96764 6.32825 6.65022 6.93770 7.19437 7.42355 7.62817 7.81086 7.97399 8.11963 8.24967 8.36578 8.46944 8.56200 8.64465 8.71843 8.78432 8.84314 8.89566 8.94255 8.98442 9.02181 9.05518 9.08499 9.11159 9.13535 9.15656 9.17550 9.19241 9.20751 9.22099 9.23303

0.125 1.00000 1.88889 2.67901 3.38134 4.00564 4.56057 5.05384 5.49230 5.88205 6.22848 6.53643 6.81016 7.05348 7.26976 7.46201 7.63289 7.78479 7.91982 8.03984 8.14652 8.24135 8.32565 8.40058 8.46718 8.52638 8.57901 8.62578 8.66736 8.70432 8.73717 8.76638 8.79234 8.81541 8.83592 8.85415 8.87036 8.88476 8.89757 8.90895 8.91906

589

A Financial Mathematical Factors qn −1 qn ·(q−1)

Annuity Value Factors (in advance) n 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

·q =

(1+i)n −1 i·(1+i)n

· (1 + i)

i 0.03 24.11477 24.41240 24.70136 24.98190 25.25427 25.51871 25.77545 26.02471 26.26671 26.50166 26.72976 26.95123 27.16624 27.37499 27.57766 27.77443 27.96546 28.15094 28.33101 28.50583 28.67556 28.84035 29.00034 29.15567 29.30648 29.45289 29.59504 29.73305 29.86704 29.99712 30.12342 30.24604 30.36509 30.48067 30.59288 30.70183 30.80760 30.91029 31.00999 31.10679

0.0375 21.55099 21.77204 21.98510 22.19046 22.38839 22.57917 22.76306 22.94030 23.11113 23.27579 23.43449 23.58746 23.73490 23.87702 24.01399 24.14602 24.27327 24.39592 24.51414 24.62809 24.73792 24.84377 24.94581 25.04415 25.13894 25.23030 25.31837 25.40324 25.48505 25.56391 25.63991 25.71317 25.78378 25.85183 25.91743 25.98065 26.04159 26.10033 26.15695 26.21151

0.04 20.79277 20.99305 21.18563 21.37079 21.54884 21.72004 21.88465 22.04294 22.19513 22.34147 22.48218 22.61749 22.74758 22.87267 22.99296 23.10861 23.21982 23.32675 23.42957 23.52843 23.62349 23.71489 23.80278 23.88729 23.96855 24.04668 24.12181 24.19405 24.26351 24.33030 24.39451 24.45626 24.51564 24.57273 24.62762 24.68041 24.73116 24.77996 24.82689 24.87201

0.0425 20.07727 20.25878 20.43288 20.59988 20.76008 20.91375 21.06115 21.20254 21.33817 21.46827 21.59306 21.71277 21.82760 21.93774 22.04340 22.14475 22.24196 22.33522 22.42467 22.51047 22.59278 22.67173 22.74746 22.82011 22.88979 22.95664 23.02075 23.08226 23.14125 23.19785 23.25213 23.30420 23.35415 23.40206 23.44802 23.49211 23.53440 23.57496 23.61387 23.65119

0.05 18.15909 18.29437 18.42321 18.54591 18.66277 18.77407 18.88007 18.98102 19.07716 19.16872 19.25593 19.33898 19.41807 19.49340 19.56515 19.63347 19.69854 19.76052 19.81954 19.87575 19.92929 19.98028 20.02883 20.07508 20.11912 20.16107 20.20102 20.23907 20.27530 20.30981 20.34268 20.37398 20.40379 20.43218 20.45922 20.48497 20.50950 20.53285 20.55510 20.57628

0.06 16.04630 16.13802 16.22454 16.30617 16.38318 16.45583 16.52437 16.58903 16.65003 16.70757 16.76186 16.81308 16.86139 16.90697 16.94998 16.99054 17.02881 17.06492 17.09898 17.13111 17.16143 17.19003 17.21701 17.24246 17.26647 17.28912 17.31049 17.33065 17.34967 17.36762 17.38454 17.40051 17.41558 17.42979 17.44320 17.45585 17.46778 17.47904 17.48966 17.49968

A Financial Mathematical Factors

590 Annuity Value Factors (in advance) n 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

qn −1 qn ·(q−1)

·q =

(1+i)n −1 i·(1+i)n

· (1 + i)

i 0.07 14.33171 14.39412 14.45245 14.50696 14.55791 14.60552 14.65002 14.69161 14.73047 14.76680 14.80075 14.83247 14.86212 14.88984 14.91573 14.93994 14.96256 14.98370 15.00346 15.02192 15.03918 15.05531 15.07038 15.08447 15.09764 15.10994 15.12144 15.13219 15.14223 15.15162 15.16039 15.16859 15.17625 15.18341 15.19010 15.19636 15.20220 15.20767 15.21277 15.21755

0.08 12.92461 12.96723 13.00670 13.04324 13.07707 13.10840 13.13741 13.16427 13.18914 13.21216 13.23348 13.25323 13.27151 13.28843 13.30410 13.31861 13.33205 13.34449 13.35601 13.36668 13.37655 13.38570 13.39416 13.40200 13.40926 13.41598 13.42221 13.42797 13.43330 13.43825 13.44282 13.44706 13.45098 13.45461 13.45797 13.46108 13.46397 13.46664 13.46911 13.47140

0.09 11.75736 11.78657 11.81337 11.83795 11.86051 11.88120 11.90018 11.91760 11.93358 11.94823 11.96168 11.97402 11.98534 11.99573 12.00525 12.01399 12.02201 12.02937 12.03612 12.04231 12.04799 12.05320 12.05798 12.06237 12.06640 12.07009 12.07347 12.07658 12.07943 12.08205 12.08445 12.08665 12.08867 12.09052 12.09222 12.09378 12.09521 12.09653 12.09773 12.09883

0.10 10.77905 10.79914 10.81740 10.83400 10.84909 10.86281 10.87528 10.88662 10.89693 10.90630 10.91481 10.92256 10.92960 10.93600 10.94182 10.94711 10.95191 10.95629 10.96026 10.96387 10.96716 10.97014 10.97286 10.97532 10.97757 10.97961 10.98146 10.98315 10.98468 10.98607 10.98734 10.98849 10.98954 10.99049 10.99135 10.99214 10.99285 10.99350 10.99409 10.99463

0.12 9.24378 9.25337 9.26194 9.26959 9.27642 9.28252 9.28796 9.29282 9.29716 9.30104 9.30450 9.30759 9.31035 9.31281 9.31501 9.31697 9.31872 9.32029 9.32169 9.32294 9.32405 9.32504 9.32593 9.32673 9.32743 9.32807 9.32863 9.32913 9.32958 9.32999 9.33034 9.33066 9.33095 9.33121 9.33143 9.33164 9.33182 9.33198 9.33213 9.33226

0.125 8.92806 8.93605 8.94316 8.94947 8.95509 8.96008 8.96451 8.96846 8.97196 8.97508 8.97785 8.98031 8.98250 8.98444 8.98617 8.98771 8.98907 8.99029 8.99137 8.99232 8.99318 8.99394 8.99461 8.99521 8.99574 8.99621 8.99663 8.99701 8.99734 8.99764 8.99790 8.99813 8.99834 8.99852 8.99869 8.99883 8.99896 8.99908 8.99918 8.99927

591

A Financial Mathematical Factors qn −1 qn ·(q−1)

Annuity Value Factors (in advance) n 85 90 95 100 105

85 90 95 100 105

(1+i)n −1 i·(1+i)n

· (1 + i)

i 0.03 31.55009 31.93248 32.26234 32.54687 32.79232

0.0375 26.45616 26.65967 26.82897 26.96981 27.08696

0.04 25.07287 25.23797 25.37367 25.48520 25.57687

0.0425 23.81619 23.95019 24.05902 24.14740 24.21917

Annuity Value Factors (in advance) n

·q =

qn −1 qn ·(q−1)

0.05 20.66801 20.73987 20.79619 20.84031 20.87488

·q =

(1+i)n −1 i·(1+i)n

0.06 17.54188 17.57342 17.59699 17.61460 17.62776

· (1 + i)

i 0.07 15.23711 15.25106 15.26101 15.26810 15.27315

0.08 13.48053 13.48675 13.49098 13.49386 13.49582

0.09 12.10313 12.10593 12.10774 12.10892 12.10969

0.10 10.99667 10.99793 10.99871 10.99920 10.99950

0.12 9.33272 9.33299 9.33314 9.33322 9.33327

0.125 8.99960 8.99978 8.99988 8.99993 8.99996

A Financial Mathematical Factors

592 Annuity Value Factors (in arrears) n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

qn −1 qn ·(q−1)

=

(1+i)n −1 i·(1+i)n

i 0.03 0.97087 1.91347 2.82861 3.71710 4.57971 5.41719 6.23028 7.01969 7.78611 8.53020 9.25262 9.95400 10.63496 11.29607 11.93794 12.56110 13.16612 13.75351 14.32380 14.87747 15.41502 15.93692 16.44361 16.93554 17.41315 17.87684 18.32703 18.76411 19.18845 19.60044 20.00043 20.38877 20.76579 21.13184 21.48722 21.83225 22.16724 22.49246 22.80822 23.11477

0.0375 0.96386 1.89287 2.78831 3.65138 4.48326 5.28507 6.05790 6.80280 7.52077 8.21279 8.87979 9.52269 10.14236 10.73962 11.31530 11.87017 12.40498 12.92046 13.41731 13.89620 14.35779 14.80269 15.23151 15.64482 16.04320 16.42718 16.79729 17.15401 17.49784 17.82925 18.14867 18.45655 18.75330 19.03933 19.31501 19.58074 19.83685 20.08371 20.32165 20.55099

0.04 0.96154 1.88609 2.77509 3.62990 4.45182 5.24214 6.00205 6.73274 7.43533 8.11090 8.76048 9.38507 9.98565 10.56312 11.11839 11.65230 12.16567 12.65930 13.13394 13.59033 14.02916 14.45112 14.85684 15.24696 15.62208 15.98277 16.32959 16.66306 16.98371 17.29203 17.58849 17.87355 18.14765 18.41120 18.66461 18.90828 19.14258 19.36786 19.58448 19.79277

0.0425 0.95923 1.87936 2.76198 3.60861 4.42073 5.19974 5.94699 6.66378 7.35135 8.01089 8.64354 9.25039 9.83251 10.39090 10.92652 11.44031 11.93315 12.40590 12.85938 13.29437 13.71162 14.11187 14.49580 14.86407 15.21734 15.55620 15.88124 16.19304 16.49213 16.77902 17.05421 17.31819 17.57141 17.81430 18.04729 18.27078 18.48516 18.69080 18.88806 19.07727

0.05 0.95238 1.85941 2.72325 3.54595 4.32948 5.07569 5.78637 6.46321 7.10782 7.72173 8.30641 8.86325 9.39357 9.89864 10.37966 10.83777 11.27407 11.68959 12.08532 12.46221 12.82115 13.16300 13.48857 13.79864 14.09394 14.37519 14.64303 14.89813 15.14107 15.37245 15.59281 15.80268 16.00255 16.19290 16.37419 16.54685 16.71129 16.86789 17.01704 17.15909

0.06 0.94340 1.83339 2.67301 3.46511 4.21236 4.91732 5.58238 6.20979 6.80169 7.36009 7.88687 8.38384 8.85268 9.29498 9.71225 10.10590 10.47726 10.82760 11.15812 11.46992 11.76408 12.04158 12.30338 12.55036 12.78336 13.00317 13.21053 13.40616 13.59072 13.76483 13.92909 14.08404 14.23023 14.36814 14.49825 14.62099 14.73678 14.84602 14.94907 15.04630

593

A Financial Mathematical Factors Annuity Value Factors (in arrears) n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

qn −1 qn ·(q−1)

=

(1+i)n −1 i·(1+i)n

i 0.07 0.93458 1.80802 2.62432 3.38721 4.10020 4.76654 5.38929 5.97130 6.51523 7.02358 7.49867 7.94269 8.35765 8.74547 9.10791 9.44665 9.76322 10.05909 10.33560 10.59401 10.83553 11.06124 11.27219 11.46933 11.65358 11.82578 11.98671 12.13711 12.27767 12.40904 12.53181 12.64656 12.75379 12.85401 12.94767 13.03521 13.11702 13.19347 13.26493 13.33171

0.08 0.92593 1.78326 2.57710 3.31213 3.99271 4.62288 5.20637 5.74664 6.24689 6.71008 7.13896 7.53608 7.90378 8.24424 8.55948 8.85137 9.12164 9.37189 9.60360 9.81815 10.01680 10.20074 10.37106 10.52876 10.67478 10.80998 10.93516 11.05108 11.15841 11.25778 11.34980 11.43500 11.51389 11.58693 11.65457 11.71719 11.77518 11.82887 11.87858 11.92461

0.09 0.91743 1.75911 2.53129 3.23972 3.88965 4.48592 5.03295 5.53482 5.99525 6.41766 6.80519 7.16073 7.48690 7.78615 8.06069 8.31256 8.54363 8.75563 8.95011 9.12855 9.29224 9.44243 9.58021 9.70661 9.82258 9.92897 10.02658 10.11613 10.19828 10.27365 10.34280 10.40624 10.46444 10.51784 10.56682 10.61176 10.65299 10.69082 10.72552 10.75736

0.10 0.90909 1.73554 2.48685 3.16987 3.79079 4.35526 4.86842 5.33493 5.75902 6.14457 6.49506 6.81369 7.10336 7.36669 7.60608 7.82371 8.02155 8.20141 8.36492 8.51356 8.64869 8.77154 8.88322 8.98474 9.07704 9.16095 9.23722 9.30657 9.36961 9.42691 9.47901 9.52638 9.56943 9.60857 9.64416 9.67651 9.70592 9.73265 9.75696 9.77905

0.12 0.89286 1.69005 2.40183 3.03735 3.60478 4.11141 4.56376 4.96764 5.32825 5.65022 5.93770 6.19437 6.42355 6.62817 6.81086 6.97399 7.11963 7.24967 7.36578 7.46944 7.56200 7.64465 7.71843 7.78432 7.84314 7.89566 7.94255 7.98442 8.02181 8.05518 8.08499 8.11159 8.13535 8.15656 8.17550 8.19241 8.20751 8.22099 8.23303 8.24378

0.125 0.88889 1.67901 2.38134 3.00564 3.56057 4.05384 4.49230 4.88205 5.22848 5.53643 5.81016 6.05348 6.26976 6.46201 6.63289 6.78479 6.91982 7.03984 7.14652 7.24135 7.32565 7.40058 7.46718 7.52638 7.57901 7.62578 7.66736 7.70432 7.73717 7.76638 7.79234 7.81541 7.83592 7.85415 7.87036 7.88476 7.89757 7.90895 7.91906 7.92806

A Financial Mathematical Factors

594 Annuity Value Factors (in arrears) n 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

qn −1 qn ·(q−1)

=

(1+i)n −1 i·(1+i)n

i 0.03 23.41240 23.70136 23.98190 24.25427 24.51871 24.77545 25.02471 25.26671 25.50166 25.72976 25.95123 26.16624 26.37499 26.57766 26.77443 26.96546 27.15094 27.33101 27.50583 27.67556 27.84035 28.00034 28.15567 28.30648 28.45289 28.59504 28.73305 28.86704 28.99712 29.12342 29.24604 29.36509 29.48067 29.59288 29.70183 29.80760 29.91029 30.00999 30.10679 30.20076

0.0375 20.77204 20.98510 21.19046 21.38839 21.57917 21.76306 21.94030 22.11113 22.27579 22.43449 22.58746 22.73490 22.87702 23.01399 23.14602 23.27327 23.39592 23.51414 23.62809 23.73792 23.84377 23.94581 24.04415 24.13894 24.23030 24.31837 24.40324 24.48505 24.56391 24.63991 24.71317 24.78378 24.85183 24.91743 24.98065 25.04159 25.10033 25.15695 25.21151 25.26411

0.04 19.99305 20.18563 20.37079 20.54884 20.72004 20.88465 21.04294 21.19513 21.34147 21.48218 21.61749 21.74758 21.87267 21.99296 22.10861 22.21982 22.32675 22.42957 22.52843 22.62349 22.71489 22.80278 22.88729 22.96855 23.04668 23.12181 23.19405 23.26351 23.33030 23.39451 23.45626 23.51564 23.57273 23.62762 23.68041 23.73116 23.77996 23.82689 23.87201 23.91539

0.0425 19.25878 19.43288 19.59988 19.76008 19.91375 20.06115 20.20254 20.33817 20.46827 20.59306 20.71277 20.82760 20.93774 21.04340 21.14475 21.24196 21.33522 21.42467 21.51047 21.59278 21.67173 21.74746 21.82011 21.88979 21.95664 22.02075 22.08226 22.14125 22.19785 22.25213 22.30420 22.35415 22.40206 22.44802 22.49211 22.53440 22.57496 22.61387 22.65119 22.68700

0.05 17.29437 17.42321 17.54591 17.66277 17.77407 17.88007 17.98102 18.07716 18.16872 18.25593 18.33898 18.41807 18.49340 18.56515 18.63347 18.69854 18.76052 18.81954 18.87575 18.92929 18.98028 19.02883 19.07508 19.11912 19.16107 19.20102 19.23907 19.27530 19.30981 19.34268 19.37398 19.40379 19.43218 19.45922 19.48497 19.5095 19.53285 19.55510 19.57628 19.59646

0.06 15.13802 15.22454 15.30617 15.38318 15.45583 15.52437 15.58903 15.65003 15.70757 15.76186 15.81308 15.86139 15.90697 15.94998 15.99054 16.02881 16.06492 16.09898 16.13111 16.16143 16.19003 16.21701 16.24246 16.26647 16.28912 16.31049 16.33065 16.34967 16.36762 16.38454 16.40051 16.41558 16.42979 16.44320 16.45585 16.46778 16.47904 16.48966 16.49968 16.50913

595

A Financial Mathematical Factors Annuity Value Factors (in arrears) n 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

qn −1 qn ·(q−1)

=

(1+i)n −1 i·(1+i)n

i 0.07 13.39412 13.45245 13.50696 13.55791 13.60552 13.65002 13.69161 13.73047 13.76680 13.80075 13.83247 13.86212 13.88984 13.91573 13.93994 13.96256 13.98370 14.00346 14.02192 14.03918 14.05531 14.07038 14.08447 14.09764 14.10994 14.12144 14.13219 14.14223 14.15162 14.16039 14.16859 14.17625 14.18341 14.19010 14.19636 14.20220 14.20767 14.21277 14.21755 14.22201

0.08 11.96723 12.00670 12.04324 12.07707 12.10840 12.13741 12.16427 12.18914 12.21216 12.23348 12.25323 12.27151 12.28843 12.30410 12.31861 12.33205 12.34449 12.35601 12.36668 12.37655 12.38570 12.39416 12.40200 12.40926 12.41598 12.42221 12.42797 12.4333 12.43825 12.44282 12.44706 12.45098 12.45461 12.45797 12.46108 12.46397 12.46664 12.46911 12.47140 12.47351

0.09 10.78657 10.81337 10.83795 10.86051 10.88120 10.90018 10.91760 10.93358 10.94823 10.96168 10.97402 10.98534 10.99573 11.00525 11.01399 11.02201 11.02937 11.03612 11.04231 11.04799 11.05320 11.05798 11.06237 11.06640 11.07009 11.07347 11.07658 11.07943 11.08205 11.08445 11.08665 11.08867 11.09052 11.09222 11.09378 11.09521 11.09653 11.09773 11.09883 11.09985

0.10 9.79914 9.81740 9.83400 9.84909 9.86281 9.87528 9.88662 9.89693 9.90630 9.91481 9.92256 9.92960 9.93600 9.94182 9.94711 9.95191 9.95629 9.96026 9.96387 9.96716 9.97014 9.97286 9.97532 9.97757 9.97961 9.98146 9.98315 9.98468 9.98607 9.98734 9.98849 9.98954 9.99049 9.99135 9.99214 9.99285 9.99350 9.99409 9.99463 9.99512

0.12 8.25337 8.26194 8.26959 8.27642 8.28252 8.28796 8.29282 8.29716 8.30104 8.30450 8.30759 8.31035 8.31281 8.31501 8.31697 8.31872 8.32029 8.32169 8.32294 8.32405 8.32504 8.32593 8.32673 8.32743 8.32807 8.32863 8.32913 8.32958 8.32999 8.33034 8.33066 8.33095 8.33121 8.33143 8.33164 8.33182 8.33198 8.33213 8.33226 8.33237

0.125 7.93605 7.94316 7.94947 7.95509 7.96008 7.96451 7.96846 7.97196 7.97508 7.97785 7.98031 7.98250 7.98444 7.98617 7.98771 7.98907 7.99029 7.99137 7.99232 7.99318 7.99394 7.99461 7.99521 7.99574 7.99621 7.99663 7.99701 7.99734 7.99764 7.99790 7.99813 7.99834 7.99852 7.99869 7.99883 7.99896 7.99908 7.99918 7.99927 7.99935

A Financial Mathematical Factors

596 Annuity Value Factors (in arrears) n 85 90 95 100 105

85 90 95 100 105

=

(1+i)n −1 i·(1+i)n

i 0.03 30.63115 31.00241 31.32266 31.59891 31.83720

0.0375 25.49991 25.69607 25.85925 25.99499 26.10792

0.04 24.10853 24.26728 24.39776 24.50500 24.59315

Annuity Value Factors (in arrears) n

qn −1 qn ·(q−1)

0.0425 22.84527 22.97381 23.07820 23.16297 23.23182

qn −1 qn ·(q−1)

=

0.05 19.68382 19.75226 19.80589 19.84791 19.88083

0.06 16.54895 16.57870 16.60093 16.61755 16.62996

(1+i)n −1 i·(1+i)n

i 0.07 14.24029 14.25333 14.26262 14.26925 14.27398

0.08 12.48197 12.48773 12.49165 12.49432 12.49613

0.09 11.10379 11.10635 11.10802 11.10910 11.10981

0.10 9.99697 9.99812 9.99883 9.99927 9.99955

0.12 8.33279 8.33302 8.33316 8.33323 8.33328

0.125 7.99964 7.99980 7.99989 7.99994 7.99997

597

A Financial Mathematical Factors Accumulation Factors of Annuity (in advance) (1+i)n −1 qn −1 · (1 + i) q−1 · q = i n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

i 0.03 1.03000 2.09090 3.18363 4.30914 5.46841 6.66246 7.89234 9.15911 10.46388 11.80780 13.19203 14.61779 16.08632 17.59891 19.15688 20.76159 22.41444 24.11687 25.87037 27.67649 29.53678 31.45288 33.42647 35.45926 37.55304 39.70963 41.93092 44.21885 46.57542 49.00268 51.50276 54.07784 56.73018 59.46208 62.27594 65.17422 68.15945 71.23423 74.40126 77.66330

0.0375 1.03750 2.11391 3.23068 4.38933 5.59143 6.83861 8.13255 9.47503 10.86784 12.31288 13.81212 15.36757 16.98135 18.65565 20.39274 22.19497 24.06478 26.00471 28.01739 30.10554 32.27200 34.51970 36.85168 39.27112 41.78129 44.38559 47.08755 49.89083 52.79924 55.81671 58.94734 62.19536 65.56519 69.06138 72.68868 76.45201 80.35646 84.40733 88.61010 92.97048

0.04 1.04000 2.12160 3.24646 4.41632 5.63298 6.89829 8.21423 9.58280 11.00611 12.48635 14.02581 15.62684 17.29191 19.02359 20.82453 22.69751 24.64541 26.67123 28.77808 30.96920 33.24797 35.61789 38.08260 40.64591 43.31174 46.08421 48.96758 51.96629 55.08494 58.32834 61.70147 65.20953 68.85791 72.65222 76.59831 80.70225 84.97034 89.40915 94.02552 98.82654

0.0425 1.04250 2.12931 3.26230 4.44345 5.67480 6.95848 8.29671 9.69182 11.14622 12.66244 14.24309 15.89092 17.60879 19.39966 21.26665 23.21298 25.24203 27.35732 29.56250 31.86141 34.25802 36.75648 39.36113 42.07648 44.90723 47.85829 50.93477 54.14199 57.48553 60.97116 64.60494 68.39315 72.34236 76.45941 80.75143 85.22587 89.89047 94.75331 99.82283 105.10780

0.05 1.05000 2.15250 3.31013 4.52563 5.80191 7.14201 8.54911 10.02656 11.57789 13.20679 14.91713 16.71298 18.59863 20.57856 22.65749 24.84037 27.13238 29.53900 32.06595 34.71925 37.50521 40.43048 43.50200 46.72710 50.11345 53.66913 57.40258 61.32271 65.43885 69.76079 74.29883 79.06377 84.06696 89.32031 94.83632 100.6281 106.7095 113.0950 119.7998 126.8398

0.06 1.06000 2.18360 3.37462 4.63709 5.97532 7.39384 8.89747 10.49132 12.18079 13.97164 15.86994 17.88214 20.01507 22.27597 24.67253 27.21288 29.90565 32.75999 35.78559 38.99273 42.39229 45.99583 49.81558 53.86451 58.15638 62.70577 67.52811 72.63980 78.05819 83.80168 89.88978 96.34316 103.1838 110.4348 118.1209 126.2681 134.9042 144.0585 153.7620 164.0477

A Financial Mathematical Factors

598

Accumulation Factors of Annuity (in advance) (1+i)n −1 qn −1 · (1 + i) q−1 · q = i n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

i 0.07 1.07000 2.21490 3.43994 4.75074 6.15329 7.65402 9.25980 10.97799 12.81645 14.78360 16.88845 19.14064 21.55049 24.12902 26.88805 29.84022 32.99903 36.37896 39.99549 43.86518 48.00574 52.43614 57.17667 62.24904 67.67647 73.48382 79.69769 86.34653 93.46079 101.0730 109.2182 117.9334 127.2588 137.2369 147.9135 159.3374 171.5610 184.6403 198.6351 213.6096

0.08 1.08000 2.24640 3.50611 4.86660 6.33593 7.92280 9.63663 11.48756 13.48656 15.64549 17.97713 20.49530 23.21492 26.15211 29.32428 32.75023 36.45024 40.44626 44.76196 49.42292 54.45676 59.89330 65.76476 72.10594 78.95442 86.35077 94.33883 102.9659 112.2832 122.3459 133.2135 144.9506 157.6267 171.3168 186.1021 202.0703 219.3159 237.9412 258.0565 279.7810

0.09 1.09000 2.27810 3.57313 4.98471 6.52333 8.20043 10.02847 12.02104 14.19293 16.56029 19.14072 21.95338 25.01919 28.36092 32.00340 35.97370 40.30134 45.01846 50.16012 55.76453 61.87334 68.53194 75.78981 83.70090 92.32398 101.7231 111.9682 123.1354 135.3075 148.5752 163.0370 178.8003 195.9823 214.7108 235.1247 257.3759 281.6298 308.0665 336.8824 368.2919

0.10 1.10000 2.31000 3.64100 5.10510 6.71561 8.48717 10.43589 12.57948 14.93742 17.53117 20.38428 23.52271 26.97498 30.77248 34.94973 39.54470 44.59917 50.15909 56.27500 63.00250 70.40275 78.54302 87.49733 97.34706 108.1818 120.0999 133.2099 147.6309 163.4940 180.9434 200.1378 221.2515 244.4767 270.0244 298.1268 329.0395 363.0434 400.4478 441.5926 486.8518

0.12 1.12000 2.37440 3.77933 5.35285 7.11519 9.08901 11.29969 13.77566 16.54874 19.65458 23.13313 27.02911 31.39260 36.27971 41.75328 47.88367 54.74971 62.43968 71.05244 80.69874 91.50258 103.6029 117.1552 132.3339 149.3339 168.3740 189.6989 213.5828 240.3327 270.2926 303.8477 341.4294 383.5210 430.6635 483.4631 542.5987 608.8305 683.0102 766.0914 859.1424

0.125 1.12500 2.39063 3.81445 5.41626 7.21829 9.24558 11.52628 14.09206 16.97857 20.22589 23.87913 27.98902 32.61264 37.81422 43.66600 50.24925 57.65541 65.98733 75.36075 85.90584 97.76908 111.1152 126.1296 143.0208 162.0234 183.4013 207.4515 234.5079 264.9464 299.1897 337.7135 381.0526 429.8092 484.6604 546.3679 615.7889 693.8875 781.7485 880.5920 991.7910

599

A Financial Mathematical Factors Accumulation Factors of Annuity (in advance) (1+i)n −1 qn −1 · (1 + i) q−1 · q = i n 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

i 0.03 81.02320 84.48389 88.04841 91.71986 95.50146 99.39650 103.4084 107.5406 111.7969 116.1808 120.6962 125.3471 130.1375 135.0716 140.1538 145.3884 150.7800 156.3334 162.0534 167.9450 174.0134 180.2638 186.7017 193.3328 200.1627 207.1976 214.4436 221.9069 229.5941 237.5119 245.6672 254.0673 262.7193 271.6309 280.8098 290.2641 300.0020 310.0321 320.3630 331.0039

0.0375 97.49437 102.1879 107.0575 112.1096 117.3512 122.7894 128.4315 134.2852 140.3584 146.6593 153.1965 159.9789 167.0156 174.3162 181.8906 189.7490 197.9020 206.3609 215.1369 224.2420 233.6886 243.4894 253.6578 264.2074 275.1527 286.5085 298.2900 310.5134 323.1952 336.3525 350.0032 364.1658 378.8595 394.1043 409.9207 426.3302 443.3551 461.0184 479.3441 498.3570

0.04 103.8196 109.0124 114.4129 120.0294 125.8706 131.9454 138.2632 144.8337 151.6671 158.7738 166.1647 173.8513 181.8454 190.1592 198.8055 207.7978 217.1497 226.8757 236.9907 247.5103 258.4507 269.8288 281.6619 293.9684 306.7671 320.0778 333.9209 348.3177 363.2905 378.8621 395.0566 411.8988 429.4148 447.6314 466.5766 486.2797 506.7709 528.0817 550.2450 573.2948

0.0425 110.6174 116.3611 122.3490 128.5913 135.0989 141.8831 148.9557 156.3288 164.0153 172.0284 180.3821 189.0909 198.1697 207.6344 217.5014 227.7877 238.5112 249.6904 261.3447 273.4944 286.1604 299.3647 313.1302 327.4808 342.4412 358.0374 374.2965 391.2466 408.9171 427.3386 446.5430 466.5636 487.4350 509.1935 531.8767 555.5240 580.1762 605.8762 632.6685 660.5994

0.05 134.2318 141.9933 150.1430 158.7002 167.6852 177.1194 187.0254 197.4267 208.3480 219.8154 231.8562 244.4990 257.7739 271.7126 286.3482 301.7157 317.8514 334.7940 352.5837 371.2629 390.8760 411.4699 433.0933 455.7980 479.6379 504.6698 530.9533 558.5510 587.5285 617.9549 649.9027 683.4478 718.6702 755.6537 794.4864 835.2607 878.0738 923.0274 970.2288 1019.790

0.06 174.9505 186.5076 198.7580 211.7435 225.5081 240.0986 255.5645 271.9584 289.3359 307.7561 327.2814 347.9783 369.9170 393.1720 417.8223 443.9517 471.6488 501.0077 532.1282 565.1159 600.0828 637.1478 676.4367 718.0829 762.2278 809.0215 858.6228 911.2002 966.9322 1026.008 1088.629 1155.006 1225.367 1299.949 1379.006 1462.806 1551.634 1645.792 1745.600 1851.396

A Financial Mathematical Factors

600

Accumulation Factors of Annuity (in advance) (1+i)n −1 qn −1 · (1 + i) q−1 · q = i n 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

i 0.07 229.6322 246.7765 265.1209 284.7493 305.7518 328.2244 352.2701 377.9990 405.5289 434.9860 466.5050 500.2303 536.3164 574.9286 616.2436 660.4506 707.7522 758.3648 812.5204 870.4668 932.4695 998.8124 1069.799 1145.755 1227.028 1313.990 1407.039 1506.602 1613.134 1727.124 1849.092 1979.599 2119.241 2268.657 2428.533 2599.601 2782.643 2978.498 3188.063 3412.297

0.08 303.2435 328.5830 355.9496 385.5056 417.4261 451.9002 489.1322 529.3427 572.7702 619.6718 670.3255 725.0316 784.1141 847.9232 916.8371 991.2640 1071.645 1158.457 1252.213 1353.470 1462.828 1580.934 1708.489 1846.248 1995.028 2155.710 2329.247 2516.667 2719.080 2937.686 3173.781 3428.764 3704.145 4001.557 4322.761 4669.662 5044.315 5448.940 5885.935 6357.890

0.09 402.5281 439.8457 480.5218 524.8587 573.1860 625.8628 683.2804 745.8656 814.0836 888.4411 969.4908 1057.835 1154.130 1259.092 1373.500 1498.205 1634.134 1782.296 1943.792 2119.823 2311.698 2520.840 2748.806 2997.288 3268.134 3563.357 3885.149 4235.902 4618.223 5034.953 5489.189 5984.306 6523.984 7112.232 7753.423 8452.321 9214.120 10044.48 10949.57 11936.13

0.10 536.6370 591.4007 651.6408 717.9048 790.7953 870.9749 959.1723 1056.190 1162.909 1280.299 1409.429 1551.472 1707.719 1879.591 2068.651 2276.616 2505.377 2757.015 3033.816 3338.298 3673.228 4041.651 4446.916 4892.707 5383.078 5922.486 6515.834 7168.518 7886.470 8676.217 9544.938 10500.53 11551.69 12707.95 13979.85 15378.93 16917.93 18610.82 20473.00 22521.40

0.12 963.3595 1080.083 1210.813 1357.230 1521.218 1704.884 1910.590 2140.981 2399.018 2688.020 3011.703 3374.227 3780.255 4235.005 4744.326 5314.765 5953.656 6669.215 7470.641 8368.238 9373.547 10499.49 11760.55 13172.94 14754.81 16526.51 18510.81 20733.22 23222.33 26010.13 29132.47 32629.48 36546.14 40932.80 45845.85 51348.48 57511.41 64413.90 72144.69 80803.18

0.125 1116.890 1257.626 1415.954 1594.074 1794.458 2019.890 2273.501 2558.814 2879.791 3240.890 3647.126 4104.142 4618.284 5196.695 5847.407 6579.458 7403.015 8329.517 9371.831 10544.44 11863.61 13347.69 15017.28 16895.56 19008.63 21385.84 24060.19 27068.84 30453.57 34261.39 38545.19 43364.47 48786.15 54885.54 61747.36 69466.91 78151.39 87921.44 98912.75 111278.0

601

A Financial Mathematical Factors Accumulation Factors of Annuity (in advance) (1+i)n −1 qn −1 · (1 + i) q−1 · q = i n 85 90 95 100 105

i 0.03 389.1927 456.6494 534.8502 625.5064 730.6017

0.0375 604.6663 732.4606 886.0822 1070.751 1292.741

0.04 703.1337 861.1027 1053.296 1287.129 1571.622

0.0425 819.1016 1014.273 1254.596 1550.518 1914.899

0.05 1307.341 1674.338 2142.728 2740.526 3503.485

0.06 2483.561 3329.540 4461.651 5976.670 8004.108

Accumulation Factors of Annuity (in advance) (1+i)n −1 qn −1 · (1 + i) q−1 · q = i n 85 90 95 100 105

i 0.07 4792.076 6727.288 9441.523 13248.38 18587.69

0.08 9348.163 13741.85 20197.63 29683.28 43620.81

0.09 18371.73 28273.71 43509.13 66950.72 103018.5

0.10 36277.66 58432.25 94112.44 151575.7 244121.0

0.12 142409.9 250982.1 442323.2 779531.8 1373809

0.125 200533.7 361375.5 651217.6 1173522.5 2114732.9

A Financial Mathematical Factors

602

Accumulation Factors of Annuity (in arrears) n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

qn −1 q−1

=

(1+i)n −1 i

i 0.03 1.0000 2.0300 3.0909 4.1836 5.3091 6.4684 7.6625 8.8923 10.1591 11.4639 12.8078 14.1920 15.6178 17.0863 18.5989 20.1569 21.7616 23.4144 25.1169 26.8704 28.6765 30.5368 32.4529 34.4265 36.4593 38.5530 40.7096 42.9309 45.2189 47.5754 50.0027 52.5028 55.0778 57.7302 60.4621 63.2759 66.1742 69.1594 72.2342 75.4013

0.0375 1.0000 2.0375 3.1139 4.2307 5.3893 6.5914 7.8386 9.1326 10.4750 11.8678 13.3129 14.8121 16.3676 17.9814 19.6557 21.3927 23.1950 25.0648 27.0047 29.0174 31.1055 33.2720 35.5197 37.8517 40.2711 42.7813 45.3856 48.0875 50.8908 53.7992 56.8167 59.9473 63.1954 66.5652 70.0614 73.6887 77.4520 81.3565 85.4073 89.6101

0.04 1.0000 2.0400 3.1216 4.2465 5.4163 6.6330 7.8983 9.2142 10.5828 12.0061 13.4864 15.0258 16.6268 18.2919 20.0236 21.8245 23.6975 25.6454 27.6712 29.7781 31.9692 34.2480 36.6179 39.0826 41.6459 44.3117 47.0842 49.9676 52.9663 56.0849 59.3283 62.7015 66.2095 69.8579 73.6522 77.5983 81.7022 85.9703 90.4091 95.0255

0.0425 1.0000 2.0425 3.1293 4.2623 5.4434 6.6748 7.9585 9.2967 10.6918 12.1462 13.6624 15.2431 16.8909 18.6088 20.3997 22.2666 24.2130 26.2420 28.3573 30.5625 32.8614 35.2580 37.7565 40.3611 43.0765 45.9072 48.8583 51.9348 55.1420 58.4855 61.9712 65.6049 69.3931 73.3424 77.4594 81.7514 86.2259 90.8905 95.7533 100.8228

0.05 1.0000 2.0500 3.1525 4.3101 5.5256 6.8019 8.1420 9.5491 11.0266 12.5779 14.2068 15.9171 17.7130 19.5986 21.5786 23.6575 25.8404 28.1324 30.5390 33.0660 35.7193 38.5052 41.4305 44.5020 47.7271 51.1135 54.6691 58.4026 62.3227 66.4388 70.7608 75.2988 80.0638 85.0670 90.3203 95.8363 101.6281 107.7095 114.0950 120.7998

0.06 1.0000 2.0600 3.1836 4.3746 5.6371 6.9753 8.3938 9.8975 11.4913 13.1808 14.9716 16.8699 18.8821 21.0151 23.2760 25.6725 28.2129 30.9057 33.7600 36.7856 39.9927 43.3923 46.9958 50.8156 54.8645 59.1564 63.7058 68.5281 73.6398 79.0582 84.8017 90.8898 97.3432 104.1838 111.4348 119.1209 127.2681 135.9042 145.0585 154.7620

603

A Financial Mathematical Factors Accumulation Factors of Annuity (in arrears) n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

qn −1 q−1

=

(1+i)n −1 i

i 0.07 1.000 2.0700 3.2149 4.4399 5.7507 7.1533 8.6540 10.2598 11.9780 13.8164 15.7836 17.8885 20.1406 22.5505 25.1290 27.8881 30.8402 33.9990 37.3790 40.9955 44.8652 49.0057 53.4361 58.1767 63.2490 68.6765 74.4838 80.6977 87.3465 94.4608 102.0730 110.2182 118.9334 128.2588 138.2369 148.9135 160.3374 172.5610 185.6403 199.6351

0.08 1.000 2.0800 3.2464 4.5061 5.8666 7.3359 8.9228 10.6366 12.4876 14.4866 16.6455 18.9771 21.4953 24.2149 27.1521 30.3243 33.7502 37.4502 41.4463 45.7620 50.4229 55.4568 60.8933 66.7648 73.1059 79.9544 87.3508 95.3388 103.9659 113.2832 123.3459 134.2135 145.9506 158.6267 172.3168 187.1021 203.0703 220.3159 238.9412 259.0565

0.09 1.000 2.0900 3.2781 4.5731 5.9847 7.5233 9.2004 11.0285 13.0210 15.1929 17.5603 20.1407 22.9534 26.0192 29.3609 33.0034 36.9737 41.3013 46.0185 51.1601 56.7645 62.8733 69.5319 76.7898 84.7009 93.3240 102.7231 112.9682 124.1354 136.3075 149.5752 164.0370 179.8003 196.9823 215.7108 236.1247 258.3759 282.6298 309.0665 337.8824

0.10 1.000 2.100 3.3100 4.6410 6.1051 7.7156 9.4872 11.4359 13.5795 15.9374 18.5312 21.3843 24.5227 27.9750 31.7725 35.9497 40.5447 45.5992 51.1591 57.2750 64.0025 71.4027 79.5430 88.4973 98.3471 109.1818 121.0999 134.2099 148.6309 164.4940 181.9434 201.1378 222.2515 245.4767 271.0244 299.1268 330.0395 364.0434 401.4478 442.5926

0.12 1.000 2.1200 3.3744 4.7793 6.3528 8.1152 10.0890 12.2997 14.7757 17.5487 20.6546 24.1331 28.0291 32.3926 37.2797 42.7533 48.8837 55.7497 63.4397 72.0524 81.6987 92.5026 104.6029 118.1552 133.3339 150.3339 169.3740 190.6989 214.5828 241.3327 271.2926 304.8477 342.4294 384.5210 431.6635 484.4631 543.5987 609.8305 684.0102 767.0914

0.125 1.000 2.1250 3.3906 4.8145 6.4163 8.2183 10.2456 12.5263 15.0921 17.9786 21.2259 24.8791 28.9890 33.6126 38.8142 44.6660 51.2493 58.6554 66.9873 76.3608 86.9058 98.7691 112.1152 127.1296 144.0208 163.0234 184.4013 208.4515 235.5079 265.9464 300.1897 338.7135 382.0526 430.8092 485.6604 547.3679 616.7889 694.8875 782.7485 881.5920

A Financial Mathematical Factors

604

Accumulation Factors of Annuity (in arrears) n 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

qn −1 q−1

=

(1+i)n −1 i

i 0.03 78.6633 82.0232 85.4839 89.0484 92.7199 96.5015 100.3965 104.4084 108.5406 112.7969 117.1808 121.6962 126.3471 131.1375 136.0716 141.1538 146.3884 151.7800 157.3334 163.0534 168.9450 175.0134 181.2638 187.7017 194.3328 201.1627 208.1976 215.4436 222.9069 230.5941 238.5119 246.6672 255.0673 263.7193 272.6309 281.8098 291.2641 301.0020 311.0321 321.3630

0.0375 93.9705 98.4944 103.1879 108.0575 113.1096 118.3512 123.7894 129.4315 135.2852 141.3584 147.6593 154.1965 160.9789 168.0156 175.3162 182.8906 190.7490 198.9020 207.3609 216.1369 225.2420 234.6886 244.4894 254.6578 265.2074 276.1527 287.5085 299.2900 311.5134 324.1952 337.3525 351.0032 365.1658 379.8595 395.1043 410.9207 427.3302 444.3551 462.0184 480.3441

0.04 99.8265 104.8196 110.0124 115.4129 121.0294 126.8706 132.9454 139.2632 145.8337 152.6671 159.7738 167.1647 174.8513 182.8454 191.1592 199.8055 208.7978 218.1497 227.8757 237.9907 248.5103 259.4507 270.8288 282.6619 294.9684 307.7671 321.0778 334.9209 349.3177 364.2905 379.8621 396.0566 412.8988 430.4148 448.6314 467.5766 487.2797 507.7709 529.0817 551.2450

0.0425 106.1078 111.6174 117.3611 123.3490 129.5913 136.0989 142.8831 149.9557 157.3288 165.0153 173.0284 181.3821 190.0909 199.1697 208.6344 218.5014 228.7877 239.5112 250.6904 262.3447 274.4944 287.1604 300.3647 314.1302 328.4808 343.4412 359.0374 375.2965 392.2466 409.9171 428.3386 447.5430 467.5636 488.4350 510.1935 532.8767 556.5240 581.1762 606.8762 633.6685

0.05 127.8398 135.2318 142.9933 151.1430 159.7002 168.6852 178.1194 188.0254 198.4267 209.3480 220.8154 232.8562 245.4990 258.7739 272.7126 287.3482 302.7157 318.8514 335.7940 353.5837 372.2629 391.8760 412.4699 434.0933 456.7980 480.6379 505.6698 531.9533 559.5510 588.5285 618.9549 650.9027 684.4478 719.6702 756.6537 795.4864 836.2607 879.0738 924.0274 971.2288

0.06 165.0477 175.9505 187.5076 199.7580 212.7435 226.5081 241.0986 256.5645 272.9584 290.3359 308.7561 328.2814 348.9783 370.9170 394.1720 418.8223 444.9517 472.6488 502.0077 533.1282 566.1159 601.0828 638.1478 677.4367 719.0829 763.2278 810.0215 859.6228 912.2002 967.9322 1027.0081 1089.6286 1156.0063 1226.3667 1300.9487 1380.0056 1463.8059 1552.6343 1646.7924 1746.5999

605

A Financial Mathematical Factors Accumulation Factors of Annuity (in arrears) n 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

qn −1 q−1

=

(1+i)n −1 i

i 0.07 214.61 230.63 247.78 266.12 285.75 306.75 329.22 353.27 379.00 406.53 435.99 467.50 501.23 537.32 575.93 617.24 661.45 708.75 759.36 813.52 871.47 933.47 999.81 1070.80 1146.76 1228.03 1314.99 1408.04 1507.60 1614.13 1728.12 1850.09 1980.60 2120.24 2269.66 2429.53 2600.60 2783.64 2979.50 3189.06

0.08 280.78 304.24 329.58 356.95 386.51 418.43 452.90 490.13 530.34 573.77 620.67 671.33 726.03 785.11 848.92 917.84 992.26 1072.65 1159.46 1253.21 1354.47 1463.83 1581.93 1709.49 1847.25 1996.03 2156.71 2330.25 2517.67 2720.08 2938.69 3174.78 3429.76 3705.15 4002.56 4323.76 4670.66 5045.32 5449.94 5886.94

0.09 369.29 403.53 440.85 481.52 525.86 574.19 626.86 684.28 746.87 815.08 889.44 970.49 1058.83 1155.13 1260.09 1374.50 1499.21 1635.13 1783.30 1944.79 2120.82 2312.70 2521.84 2749.81 2998.29 3269.13 3564.36 3886.15 4236.90 4619.22 5035.95 5490.19 5985.31 6524.98 7113.23 7754.42 8453.32 9215.12 10045.48 10950.57

0.10 487.85 537.64 592.40 652.64 718.90 791.80 871.97 960.17 1057.19 1163.91 1281.30 1410.43 1552.47 1708.72 1880.59 2069.65 2277.62 2506.38 2758.01 3034.82 3339.30 3674.23 4042.65 4447.92 4893.71 5384.08 5923.49 6516.83 7169.52 7887.47 8677.22 9545.94 10501.53 11552.69 12708.95 13980.85 15379.93 16918.93 18611.82 20474.00

0.12 860.14 964.36 1081.08 1211.81 1358.23 1522.22 1705.88 1911.59 2141.98 2400.02 2689.02 3012.70 3375.23 3781.25 4236.01 4745.33 5315.76 5954.66 6670.22 7471.64 8369.24 9374.55 10500.49 11761.55 13173.94 14755.81 16527.51 18511.81 20734.22 23223.33 26011.13 29133.47 32630.48 36547.14 40933.80 45846.85 51349.48 57512.41 64414.90 72145.69

0.125 992.79 1117.89 1258.63 1416.95 1595.07 1795.46 2020.89 2274.50 2559.81 2880.79 3241.89 3648.13 4105.14 4619.28 5197.70 5848.41 6580.46 7404.01 8330.52 9372.83 10545.44 11864.61 13348.69 15018.28 16896.56 19009.63 21386.84 24061.19 27069.84 30454.57 34262.39 38546.19 43365.47 48787.15 54886.54 61748.36 69467.91 78152.39 87922.44 98913.75

A Financial Mathematical Factors

606

Accumulation Factors of Annuity (in arrears) n 85 90 95 100 105

85 90 95 100 105

=

(1+i)n −1 i

i 0.03 377.8570 443.3489 519.2720 607.2877 709.3221

0.0375 582.8109 705.9861 854.0551 1032.0488 1246.0150

0.04 676.0901 827.9833 1012.7846 1237.6237 1511.1748

0.0425 785.7090 972.9235 1203.4496 1487.3070 1836.8338

Accumulation Factors of Annuity (in arrears) n

qn −1 q−1

0.05 1245.0871 1594.6073 2040.6935 2610.0252 3336.6526

qn −1 q−1

=

0.06 2342.9817 3141.0752 4209.1042 5638.3681 7551.0454

(1+i)n −1 i

i 0.07 4478.58 6287.19 8823.85 12381.66 17371.67

0.08 8655.71 12723.94 18701.51 27484.52 40389.64

0.09 16854.80 25939.18 39916.63 61422.68 94512.38

0.10 0.12 0.125 32979.69 127151.71 178252.20 53120.23 224091.12 321222.67 85556.76 394931.47 578860.10 137796.12 696010.55 1043131.12 221928.14 1226614.75 1879762.56

607

A Financial Mathematical Factors Annuity Factors (in advance) n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

qn (q−1) qn −1

· qn =

i·(1+i)n (1+i)n −1

1 · (1+i) n

i 0.03 1.00000 0.49261 0.32353 0.23903 0.18835 0.15460 0.13051 0.11246 0.09843 0.08723 0.07808 0.07046 0.06403 0.05853 0.05377 0.04961 0.04595 0.04271 0.03981 0.03722 0.03487 0.03275 0.03081 0.02905 0.02743 0.02594 0.02456 0.02329 0.02211 0.02102 0.02000 0.01905 0.01816 0.01732 0.01654 0.01580 0.01511 0.01446 0.01384 0.01326

0.0375 1.00000 0.49080 0.32114 0.23637 0.18555 0.15171 0.12757 0.10950 0.09547 0.08426 0.07512 0.06751 0.06110 0.05561 0.05088 0.04674 0.04311 0.03990 0.03703 0.03446 0.03215 0.03006 0.02815 0.02642 0.02483 0.02337 0.02203 0.02080 0.01965 0.01859 0.01760 0.01668 0.01582 0.01502 0.01427 0.01357 0.01291 0.01229 0.01171 0.01116

0.04 1.00000 0.49020 0.32035 0.23549 0.18463 0.15076 0.12661 0.10853 0.09449 0.08329 0.07415 0.06655 0.06014 0.05467 0.04994 0.04582 0.04220 0.03899 0.03614 0.03358 0.03128 0.02920 0.02731 0.02559 0.02401 0.02257 0.02124 0.02001 0.01888 0.01783 0.01686 0.01595 0.01510 0.01431 0.01358 0.01289 0.01224 0.01163 0.01106 0.01052

0.0425 1.00000 0.48960 0.31956 0.23462 0.18371 0.14982 0.12565 0.10756 0.09353 0.08233 0.07319 0.06560 0.05920 0.05374 0.04902 0.04491 0.04130 0.03811 0.03526 0.03272 0.03043 0.02836 0.02649 0.02478 0.02321 0.02178 0.02047 0.01925 0.01813 0.01710 0.01614 0.01524 0.01441 0.01363 0.01291 0.01223 0.01160 0.01100 0.01044 0.00992

0.05 1.00000 0.48780 0.31721 0.23201 0.18097 0.14702 0.12282 0.10472 0.09069 0.07950 0.07039 0.06283 0.05646 0.05102 0.04634 0.04227 0.03870 0.03555 0.03275 0.03024 0.02800 0.02597 0.02414 0.02247 0.02095 0.01956 0.01829 0.01712 0.01605 0.01505 0.01413 0.01328 0.01249 0.01176 0.01107 0.01043 0.00984 0.00928 0.00876 0.00828

0.06 1.00000 0.48544 0.31411 0.22859 0.17740 0.14336 0.11914 0.10104 0.08702 0.07587 0.06679 0.05928 0.05296 0.04758 0.04296 0.03895 0.03544 0.03236 0.02962 0.02718 0.02500 0.02305 0.02128 0.01968 0.01823 0.01690 0.01570 0.01459 0.01358 0.01265 0.01179 0.01100 0.01027 0.00960 0.00897 0.00839 0.00786 0.00736 0.00689 0.00646

A Financial Mathematical Factors

608 Annuity Factors (in advance) n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

qn (q−1) qn −1

· qn =

i·(1+i)n (1+i)n −1

1 · (1+i) n

i 0.07 1.00000 0.48309 0.31105 0.22523 0.17389 0.13980 0.11555 0.09747 0.08349 0.07238 0.06336 0.05590 0.04965 0.04434 0.03979 0.03586 0.03243 0.02941 0.02675 0.02439 0.02229 0.02041 0.01871 0.01719 0.01581 0.01456 0.01343 0.01239 0.01145 0.01059 0.00980 0.00907 0.00841 0.00780 0.00723 0.00672 0.00624 0.00580 0.00539 0.00501

0.08 1.00000 0.48077 0.30803 0.22192 0.17046 0.13632 0.11207 0.09401 0.08008 0.06903 0.06008 0.05270 0.04652 0.04130 0.03683 0.03298 0.02963 0.02670 0.02413 0.02185 0.01983 0.01803 0.01642 0.01498 0.01368 0.01251 0.01145 0.01049 0.00962 0.00883 0.00811 0.00745 0.00685 0.00630 0.00580 0.00534 0.00492 0.00454 0.00419 0.00386

0.09 1.00000 0.47847 0.30505 0.21867 0.16709 0.13292 0.10869 0.09067 0.07680 0.06582 0.05695 0.04965 0.04357 0.03843 0.03406 0.03030 0.02705 0.02421 0.02173 0.01955 0.01762 0.01590 0.01438 0.01302 0.01181 0.01072 0.00973 0.00885 0.00806 0.00734 0.00669 0.00610 0.00556 0.00508 0.00464 0.00424 0.00387 0.00354 0.00324 0.00296

0.10 1.00000 0.47619 0.30211 0.21547 0.16380 0.12961 0.10541 0.08744 0.07364 0.06275 0.05396 0.04676 0.04078 0.03575 0.03147 0.02782 0.02466 0.02193 0.01955 0.01746 0.01562 0.01401 0.01257 0.01130 0.01017 0.00916 0.00826 0.00745 0.00673 0.00608 0.00550 0.00497 0.00450 0.00407 0.00369 0.00334 0.00303 0.00275 0.00249 0.00226

0.12 1.00000 0.47170 0.29635 0.20923 0.15741 0.12323 0.09912 0.08130 0.06768 0.05698 0.04842 0.04144 0.03568 0.03087 0.02682 0.02339 0.02046 0.01794 0.01576 0.01388 0.01224 0.01081 0.00956 0.00846 0.00750 0.00665 0.00590 0.00524 0.00466 0.00414 0.00369 0.00328 0.00292 0.00260 0.00232 0.00206 0.00184 0.00164 0.00146 0.00130

0.125 1.00000 0.47059 0.29493 0.20771 0.15585 0.12168 0.09760 0.07983 0.06626 0.05562 0.04711 0.04019 0.03450 0.02975 0.02576 0.02239 0.01951 0.01705 0.01493 0.01310 0.01151 0.01012 0.00892 0.00787 0.00694 0.00613 0.00542 0.00480 0.00425 0.00376 0.00333 0.00295 0.00262 0.00232 0.00206 0.00183 0.00162 0.00144 0.00128 0.00113

609

A Financial Mathematical Factors Annuity Factors (in advance) n 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

qn (q−1) qn −1

· qn =

i·(1+i)n (1+i)n −1

1 · (1+i) n

i 0.03 0.01271 0.01219 0.01170 0.01123 0.01079 0.01036 0.00996 0.00958 0.00921 0.00887 0.00853 0.00822 0.00791 0.00763 0.00735 0.00708 0.00683 0.00659 0.00636 0.00613 0.00592 0.00571 0.00552 0.00533 0.00515 0.00497 0.00480 0.00464 0.00449 0.00434 0.00419 0.00405 0.00392 0.00379 0.00367 0.00355 0.00343 0.00332 0.00322 0.00311

0.0375 0.01064 0.01015 0.00969 0.00925 0.00884 0.00845 0.00808 0.00773 0.00739 0.00707 0.00677 0.00649 0.00621 0.00595 0.00570 0.00547 0.00524 0.00503 0.00482 0.00463 0.00444 0.00426 0.00409 0.00393 0.00377 0.00362 0.00348 0.00334 0.00321 0.00308 0.00296 0.00285 0.00274 0.00263 0.00253 0.00243 0.00234 0.00225 0.00216 0.00208

0.04 0.01002 0.00954 0.00909 0.00866 0.00826 0.00788 0.00752 0.00718 0.00686 0.00655 0.00626 0.00598 0.00572 0.00547 0.00523 0.00500 0.00479 0.00458 0.00439 0.00420 0.00402 0.00385 0.00369 0.00354 0.00339 0.00325 0.00311 0.00299 0.00286 0.00275 0.00263 0.00252 0.00242 0.00232 0.00223 0.00214 0.00205 0.00197 0.00189 0.00181

0.0425 0.00942 0.00896 0.00852 0.00811 0.00772 0.00735 0.00700 0.00667 0.00636 0.00606 0.00578 0.00551 0.00526 0.00502 0.00479 0.00458 0.00437 0.00418 0.00399 0.00381 0.00364 0.00348 0.00333 0.00318 0.00304 0.00291 0.00279 0.00266 0.00255 0.00244 0.00233 0.00223 0.00214 0.00205 0.00196 0.00188 0.00180 0.00172 0.00165 0.00158

0.05 0.00782 0.00739 0.00699 0.00662 0.00626 0.00593 0.00561 0.00532 0.00504 0.00478 0.00453 0.00429 0.00407 0.00386 0.00367 0.00348 0.00330 0.00314 0.00298 0.00283 0.00269 0.00255 0.00242 0.00230 0.00219 0.00208 0.00198 0.00188 0.00179 0.00170 0.00162 0.00154 0.00146 0.00139 0.00132 0.00126 0.00120 0.00114 0.00108 0.00103

0.06 0.00606 0.00568 0.00533 0.00501 0.00470 0.00441 0.00415 0.00390 0.00366 0.00344 0.00324 0.00305 0.00287 0.00270 0.00254 0.00239 0.00225 0.00212 0.00199 0.00188 0.00177 0.00166 0.00157 0.00148 0.00139 0.00131 0.00123 0.00116 0.00110 0.00103 0.00097 0.00092 0.00087 0.00082 0.00077 0.00072 0.00068 0.00064 0.00061 0.00057

A Financial Mathematical Factors

610 Annuity Factors (in advance) n 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

qn (q−1) qn −1

· qn =

i·(1+i)n (1+i)n −1

1 · (1+i) n

i 0.07 0.00466 0.00434 0.00404 0.00376 0.00350 0.00326 0.00304 0.00283 0.00264 0.00246 0.00229 0.00214 0.00200 0.00186 0.00174 0.00162 0.00151 0.00141 0.00132 0.00123 0.00115 0.00107 0.00100 0.00093 0.00087 0.00081 0.00076 0.00071 0.00066 0.00062 0.00058 0.00054 0.00050 0.00047 0.00044 0.00041 0.00038 0.00036 0.00034 0.00031

0.08 0.00356 0.00329 0.00303 0.00280 0.00259 0.00239 0.00221 0.00204 0.00189 0.00174 0.00161 0.00149 0.00138 0.00127 0.00118 0.00109 0.00101 0.00093 0.00086 0.00080 0.00074 0.00068 0.00063 0.00058 0.00054 0.00050 0.00046 0.00043 0.00040 0.00037 0.00034 0.00031 0.00029 0.00027 0.00025 0.00023 0.00021 0.00020 0.00018 0.00017

0.09 0.00271 0.00248 0.00227 0.00208 0.00190 0.00174 0.00160 0.00146 0.00134 0.00123 0.00112 0.00103 0.00094 0.00087 0.00079 0.00073 0.00067 0.00061 0.00056 0.00051 0.00047 0.00043 0.00040 0.00036 0.00033 0.00031 0.00028 0.00026 0.00024 0.00022 0.00020 0.00018 0.00017 0.00015 0.00014 0.00013 0.00012 0.00011 0.00010 0.00009

0.10 0.00205 0.00186 0.00169 0.00153 0.00139 0.00126 0.00115 0.00104 0.00095 0.00086 0.00078 0.00071 0.00064 0.00059 0.00053 0.00048 0.00044 0.00040 0.00036 0.00033 0.00030 0.00027 0.00025 0.00022 0.00020 0.00019 0.00017 0.00015 0.00014 0.00013 0.00012 0.00010 0.00010 0.00009 0.00008 0.00007 0.00007 0.00006 0.00005 0.00005

0.12 0.00116 0.00104 0.00092 0.00083 0.00074 0.00066 0.00059 0.00052 0.00047 0.00042 0.00037 0.00033 0.00030 0.00026 0.00024 0.00021 0.00019 0.00017 0.00015 0.00013 0.00012 0.00011 0.00010 0.00009 0.00008 0.00007 0.00006 0.00005 0.00005 0.00004 0.00004 0.00003 0.00003 0.00003 0.00002 0.00002 0.00002 0.00002 0.00002 0.00001

0.125 0.00101 0.00089 0.00079 0.00071 0.00063 0.00056 0.00049 0.00044 0.00039 0.00035 0.00031 0.00027 0.00024 0.00022 0.00019 0.00017 0.00015 0.00014 0.00012 0.00011 0.00009 0.00008 0.00007 0.00007 0.00006 0.00005 0.00005 0.00004 0.00004 0.00003 0.00003 0.00003 0.00002 0.00002 0.00002 0.00002 0.00001 0.00001 0.00001 0.00001

611

A Financial Mathematical Factors Annuity Factors (in advance) n 85 90 95 100 105

85 90 95 100 105

· qn =

i·(1+i)n (1+i)n −1

1 · (1+i) n

i 0.03 0.00265 0.00226 0.00193 0.00165 0.00141

0.0375 0.00172 0.00142 0.00117 0.00097 0.00080

0.04 0.00148 0.00121 0.00099 0.00081 0.00066

Annuity Factors (in advance) n

qn (q−1) qn −1

0.0425 0.00127 0.00103 0.00083 0.00067 0.00054

qn (q−1) qn −1

· qn =

0.05 0.00080 0.00063 0.00049 0.00038 0.00030

i·(1+i)n (1+i)n −1

0.06 0.00043 0.00032 0.00024 0.00018 0.00013

1 · (1+i) n

i 0.07 0.00022 0.00016 0.00011 0.00008 0.00006

0.08 0.00012 0.00008 0.00005 0.00004 0.00002

0.09 0.00006 0.00004 0.00003 0.00002 0.00001

0.10 0.00003 0.00002 0.00001 0.00001 0.00000

0.12 0.00001 0.00000 0.00000 0.00000 0.00000

0.125 0.00001 0.00000 0.00000 0.00000 0.00000

A Financial Mathematical Factors

612 Annuity Factors (in arrears) n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

qn (q−1) qn −1

=

i·(1+i)n (1+i)n −1

i 0.03 1.03000 0.52261 0.35353 0.26903 0.21835 0.18460 0.16051 0.14246 0.12843 0.11723 0.10808 0.10046 0.09403 0.08853 0.08377 0.07961 0.07595 0.07271 0.06981 0.06722 0.06487 0.06275 0.06081 0.05905 0.05743 0.05594 0.05456 0.05329 0.05211 0.05102 0.05000 0.04905 0.04816 0.04732 0.04654 0.04580 0.04511 0.04446 0.04384 0.04326

0.0375 1.03750 0.52830 0.35864 0.27387 0.22305 0.18921 0.16507 0.14700 0.13297 0.12176 0.11262 0.10501 0.09860 0.09311 0.08838 0.08424 0.08061 0.07740 0.07453 0.07196 0.06965 0.06756 0.06565 0.06392 0.06233 0.06087 0.05953 0.05830 0.05715 0.05609 0.05510 0.05418 0.05332 0.05252 0.05177 0.05107 0.05041 0.04979 0.04921 0.04866

0.04 1.04000 0.53020 0.36035 0.27549 0.22463 0.19076 0.16661 0.14853 0.13449 0.12329 0.11415 0.10655 0.10014 0.09467 0.08994 0.08582 0.08220 0.07899 0.07614 0.07358 0.07128 0.06920 0.06731 0.06559 0.06401 0.06257 0.06124 0.06001 0.05888 0.05783 0.05686 0.05595 0.05510 0.05431 0.05358 0.05289 0.05224 0.05163 0.05106 0.05052

0.0425 1.04250 0.53210 0.36206 0.27712 0.22621 0.19232 0.16815 0.15006 0.13603 0.12483 0.11569 0.10810 0.10170 0.09624 0.09152 0.08741 0.08380 0.08061 0.07776 0.07522 0.07293 0.07086 0.06899 0.06728 0.06571 0.06428 0.06297 0.06175 0.06063 0.05960 0.05864 0.05774 0.05691 0.05613 0.05541 0.05473 0.05410 0.05350 0.05294 0.05242

0.05 1.05000 0.53780 0.36721 0.28201 0.23097 0.19702 0.17282 0.15472 0.14069 0.12950 0.12039 0.11283 0.10646 0.10102 0.09634 0.09227 0.08870 0.08555 0.08275 0.08024 0.07800 0.07597 0.07414 0.07247 0.07095 0.06956 0.06829 0.06712 0.06605 0.06505 0.06413 0.06328 0.06249 0.06176 0.06107 0.06043 0.05984 0.05928 0.05876 0.05828

0.06 1.06000 0.54544 0.37411 0.28859 0.23740 0.20336 0.17914 0.16104 0.14702 0.13587 0.12679 0.11928 0.11296 0.10758 0.10296 0.09895 0.09544 0.09236 0.08962 0.08718 0.08500 0.08305 0.08128 0.07968 0.07823 0.07690 0.07570 0.07459 0.07358 0.07265 0.07179 0.07100 0.07027 0.06960 0.06897 0.06839 0.06786 0.06736 0.06689 0.06646

613

A Financial Mathematical Factors Annuity Factors (in arrears) n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

qn (q−1) qn −1

=

i·(1+i)n (1+i)n −1

i 0.07 1.07000 0.55309 0.38105 0.29523 0.24389 0.20980 0.18555 0.16747 0.15349 0.14238 0.13336 0.12590 0.11965 0.11434 0.10979 0.10586 0.10243 0.09941 0.09675 0.09439 0.09229 0.09041 0.08871 0.08719 0.08581 0.08456 0.08343 0.08239 0.08145 0.08059 0.07980 0.07907 0.07841 0.07780 0.07723 0.07672 0.07624 0.07580 0.07539 0.07501

0.08 1.08000 0.56077 0.38803 0.30192 0.25046 0.21632 0.19207 0.17401 0.16008 0.14903 0.14008 0.13270 0.12652 0.12130 0.11683 0.11298 0.10963 0.10670 0.10413 0.10185 0.09983 0.09803 0.09642 0.09498 0.09368 0.09251 0.09145 0.09049 0.08962 0.08883 0.08811 0.08745 0.08685 0.08630 0.08580 0.08534 0.08492 0.08454 0.08419 0.08386

0.09 1.09000 0.56847 0.39505 0.30867 0.25709 0.22292 0.19869 0.18067 0.16680 0.15582 0.14695 0.13965 0.13357 0.12843 0.12406 0.12030 0.11705 0.11421 0.11173 0.10955 0.10762 0.10590 0.10438 0.10302 0.10181 0.10072 0.09973 0.09885 0.09806 0.09734 0.09669 0.09610 0.09556 0.09508 0.09464 0.09424 0.09387 0.09354 0.09324 0.09296

0.10 1.10000 0.57619 0.40211 0.31547 0.26380 0.22961 0.20541 0.18744 0.17364 0.16275 0.15396 0.14676 0.14078 0.13575 0.13147 0.12782 0.12466 0.12193 0.11955 0.11746 0.11562 0.11401 0.11257 0.11130 0.11017 0.10916 0.10826 0.10745 0.10673 0.10608 0.10550 0.10497 0.10450 0.10407 0.10369 0.10334 0.10303 0.10275 0.10249 0.10226

0.12 1.12000 0.59170 0.41635 0.32923 0.27741 0.24323 0.21912 0.20130 0.18768 0.17698 0.16842 0.16144 0.15568 0.15087 0.14682 0.14339 0.14046 0.13794 0.13576 0.13388 0.13224 0.13081 0.12956 0.12846 0.12750 0.12665 0.12590 0.12524 0.12466 0.12414 0.12369 0.12328 0.12292 0.12260 0.12232 0.12206 0.12184 0.12164 0.12146 0.12130

0.125 1.12500 0.59559 0.41993 0.33271 0.28085 0.24668 0.22260 0.20483 0.19126 0.18062 0.17211 0.16519 0.15950 0.15475 0.15076 0.14739 0.14451 0.14205 0.13993 0.13810 0.13651 0.13512 0.13392 0.13287 0.13194 0.13113 0.13042 0.12980 0.12925 0.12876 0.12833 0.12795 0.12762 0.12732 0.12706 0.12683 0.12662 0.12644 0.12628 0.12613

A Financial Mathematical Factors

614 Annuity Factors (in arrears) n 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

qn (q−1) qn −1

=

i·(1+i)n (1+i)n −1

i 0.03 0.04271 0.04219 0.04170 0.04123 0.04079 0.04036 0.03996 0.03958 0.03921 0.03887 0.03853 0.03822 0.03791 0.03763 0.03735 0.03708 0.03683 0.03659 0.03636 0.03613 0.03592 0.03571 0.03552 0.03533 0.03515 0.03497 0.03480 0.03464 0.03449 0.03434 0.03419 0.03405 0.03392 0.03379 0.03367 0.03355 0.03343 0.03332 0.03322 0.03311

0.0375 0.04814 0.04765 0.04719 0.04675 0.04634 0.04595 0.04558 0.04523 0.04489 0.04457 0.04427 0.04399 0.04371 0.04345 0.04320 0.04297 0.04274 0.04253 0.04232 0.04213 0.04194 0.04176 0.04159 0.04143 0.04127 0.04112 0.04098 0.04084 0.04071 0.04058 0.04046 0.04035 0.04024 0.04013 0.04003 0.03993 0.03984 0.03975 0.03966 0.03958

0.04 0.05002 0.04954 0.04909 0.04866 0.04826 0.04788 0.04752 0.04718 0.04686 0.04655 0.04626 0.04598 0.04572 0.04547 0.04523 0.04500 0.04479 0.04458 0.04439 0.04420 0.04402 0.04385 0.04369 0.04354 0.04339 0.04325 0.04311 0.04299 0.04286 0.04275 0.04263 0.04252 0.04242 0.04232 0.04223 0.04214 0.04205 0.04197 0.04189 0.04181

0.0425 0.05192 0.05146 0.05102 0.05061 0.05022 0.04985 0.04950 0.04917 0.04886 0.04856 0.04828 0.04801 0.04776 0.04752 0.04729 0.04708 0.04687 0.04668 0.04649 0.04631 0.04614 0.04598 0.04583 0.04568 0.04554 0.04541 0.04529 0.04516 0.04505 0.04494 0.04483 0.04473 0.04464 0.04455 0.04446 0.04438 0.04430 0.04422 0.04415 0.04408

0.05 0.05782 0.05739 0.05699 0.05662 0.05626 0.05593 0.05561 0.05532 0.05504 0.05478 0.05453 0.05429 0.05407 0.05386 0.05367 0.05348 0.05330 0.05314 0.05298 0.05283 0.05269 0.05255 0.05242 0.05230 0.05219 0.05208 0.05198 0.05188 0.05179 0.05170 0.05162 0.05154 0.05146 0.05139 0.05132 0.05126 0.05120 0.05114 0.05108 0.05103

0.06 0.06606 0.06568 0.06533 0.06501 0.06470 0.06441 0.06415 0.06390 0.06366 0.06344 0.06324 0.06305 0.06287 0.06270 0.06254 0.06239 0.06225 0.06212 0.06199 0.06188 0.06177 0.06166 0.06157 0.06148 0.06139 0.06131 0.06123 0.06116 0.06110 0.06103 0.06097 0.06092 0.06087 0.06082 0.06077 0.06072 0.06068 0.06064 0.06061 0.06057

615

A Financial Mathematical Factors Annuity Factors (in arrears) n 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

qn (q−1) qn −1

=

i·(1+i)n (1+i)n −1

i 0.07 0.07466 0.07434 0.07404 0.07376 0.07350 0.07326 0.07304 0.07283 0.07264 0.07246 0.07229 0.07214 0.07200 0.07186 0.07174 0.07162 0.07151 0.07141 0.07132 0.07123 0.07115 0.07107 0.07100 0.07093 0.07087 0.07081 0.07076 0.07071 0.07066 0.07062 0.07058 0.07054 0.07050 0.07047 0.07044 0.07041 0.07038 0.07036 0.07034 0.07031

0.08 0.08356 0.08329 0.08303 0.08280 0.08259 0.08239 0.08221 0.08204 0.08189 0.08174 0.08161 0.08149 0.08138 0.08127 0.08118 0.08109 0.08101 0.08093 0.08086 0.08080 0.08074 0.08068 0.08063 0.08058 0.08054 0.08050 0.08046 0.08043 0.08040 0.08037 0.08034 0.08031 0.08029 0.08027 0.08025 0.08023 0.08021 0.08020 0.08018 0.08017

0.09 0.09271 0.09248 0.09227 0.09208 0.09190 0.09174 0.09160 0.09146 0.09134 0.09123 0.09112 0.09103 0.09094 0.09087 0.09079 0.09073 0.09067 0.09061 0.09056 0.09051 0.09047 0.09043 0.09040 0.09036 0.09033 0.09031 0.09028 0.09026 0.09024 0.09022 0.09020 0.09018 0.09017 0.09015 0.09014 0.09013 0.09012 0.09011 0.09010 0.09009

0.10 0.10205 0.10186 0.10169 0.10153 0.10139 0.10126 0.10115 0.10104 0.10095 0.10086 0.10078 0.10071 0.10064 0.10059 0.10053 0.10048 0.10044 0.10040 0.10036 0.10033 0.10030 0.10027 0.10025 0.10022 0.10020 0.10019 0.10017 0.10015 0.10014 0.10013 0.10012 0.10010 0.10010 0.10009 0.10008 0.10007 0.10007 0.10006 0.10005 0.10005

0.12 0.12116 0.12104 0.12092 0.12083 0.12074 0.12066 0.12059 0.12052 0.12047 0.12042 0.12037 0.12033 0.12030 0.12026 0.12024 0.12021 0.12019 0.12017 0.12015 0.12013 0.12012 0.12011 0.12010 0.12009 0.12008 0.12007 0.12006 0.12005 0.12005 0.12004 0.12004 0.12003 0.12003 0.12003 0.12002 0.12002 0.12002 0.12002 0.12002 0.12001

0.125 0.12601 0.12589 0.12579 0.12571 0.12563 0.12556 0.12549 0.12544 0.12539 0.12535 0.12531 0.12527 0.12524 0.12522 0.12519 0.12517 0.12515 0.12514 0.12512 0.12511 0.12509 0.12508 0.12507 0.12507 0.12506 0.12505 0.12505 0.12504 0.12504 0.12503 0.12503 0.12503 0.12502 0.12502 0.12502 0.12502 0.12501 0.12501 0.12501 0.12501

A Financial Mathematical Factors

616 Annuity Factors (in arrears) n 85 90 95 100 105

85 90 95 100 105

=

i·(1+i)n (1+i)n −1

i 0.03 0.03265 0.03226 0.03193 0.03165 0.03141

0.0375 0.03922 0.03892 0.03867 0.03847 0.03830

0.04 0.04148 0.04121 0.04099 0.04081 0.04066

Annuity Factors (in arrears) n

qn (q−1) qn −1

0.0425 0.04377 0.04353 0.04333 0.04317 0.04304

qn (q−1) qn −1

=

0.05 0.05080 0.05063 0.05049 0.05038 0.05030

0.06 0.06043 0.06032 0.06024 0.06018 0.06013

i·(1+i)n (1+i)n −1

i 0.07 0.07022 0.07016 0.07011 0.07008 0.07006

0.08 0.08012 0.08008 0.08005 0.08004 0.08002

0.09 0.09006 0.09004 0.09003 0.09002 0.09001

0.10 0.10003 0.10002 0.10001 0.10001 0.10000

0.12 0.12001 0.12000 0.12000 0.12000 0.12000

0.125 0.12501 0.12500 0.12500 0.12500 0.12500

Appendix B

Bibliography

© Springer-Verlag GmbH Germany, part of Springer Nature 2023 F. W. Peren, Math for Business and Economics, https://doi.org/10.1007/978-3-662-66975-4

617

618

B Bibliography

[1] Bartsch, H.-J. (2004): Taschenbuch mathematischer Formeln, 20th Edition, Munich & Vienna, 978-3-446-22891-7. [2] Bartsch, H.-J. & Sachs, M. (2018): Taschenbuch mathematischer Formeln für Ingenieure und Naturwissenschaftler, 24th Edition, Munich, ISBN 978-3-446-45707-2. [3] Behrens, C. & Peren, F.W. (1998): Grundzuege der gesamtwirtschaftlichen Produktionstheorie, Munich, ISBN 3-8006-2198-3. [4] Brandes, J. (2019): Eigenschaften trigonometrischer Funktionen, http://elsenaju.info/Rechnen/Trigonometrie-Tabellen.htm, accessed 9 December 2022. [5] Buecker, R. (2003): Mathematik für Wirtschaftswissenschafter, 6th Edition, Berlin & Boston, ISBN 978-3-486-99389-9. [6] Domschke, W.; Drexl, A; Klein, R. & Scholl, A. (2015): Einführung in Operations Research, 9th Edition, Berlin & Heidelberg. [7] Federal Deposit Insurance Corporation (Ed.) (2009): https://www.fdic.gov/regulations/laws/rules/6500-1650.html# 6500226.14, accessed 9 December 2022. [8] Federal Deposit Insurance Corporation (Ed.) (2014): https://www.fdic.gov/regulations/laws/rules/6500-3550.html, accessed 9 December 2022. [9] Fußy, D.; Hanner, A.; Thomas-Frank, N. & Ulucan, D.M. (2021): Linear Optimization. Basics of the Simplex Method and its Relevance to Business Management, Sankt Augustin. [10] Geyer, H. (2014): Kennzahlen für die Bau- und Immobilienwirtschaft, 1st Edition, Freiburg. [11] Global Footprint Network (2017): http://www.footprintnetwork.org/, accessed 9 December 2022.

B Bibliography

619

[12] Gritzmann, P. (2013): Grundlagen der Mathematischen Optimierung, Wiesbaden. [13] Hemmerich, W. (2019): Formelsammlung Trigonometrie, https://matheguru.com/allgemein/formelsammlung-trigonometrie.html, accessed 7 December 2022. [14] Hempel, T.(2019): Mathematische Grundlagen, hyperbolische Funktionen, http://www.uni-magdeburg.de/exph/mathe_gl/hyperbelfunktion.pdf, accessed 19 July 2019. [15] Hempel, T. (2019): Mathematische Grundlagen, trigonometrische Funktionen, http://www.uni-magdeburg.de/exph/mathe_gl/ trigonometrische_funktionen.pdf, accessed 7 December 2022. [16] http://www.falk-net.de/ingo/fhma/mathematik/docus/ma_ formelsammlung.pdf, accessed 19 July 2019. [17] Karlsruher Institut für Technologie (Ed.) (2014): Trigonometrische und hyperbolische Funktionen, http://www.math.kit.edu/iana3/lehre/hm2inf2014s/media/ trigonometrische_hyperbolische_funktionen.pdf, accessed 19 July 2019. [18] Koop, A. & Moock, H. (2018): Lineare Optimierung – eine anwendungsorientierte Einführung in Operations Research, 2nd Edition, Berlin. [19] International Centre for Sustainable Development – IZNE (2017): https://www.h-brs.de/en/izne, accessed 9 December 2022. [20] Kruschwitz, L. (2010): Finanzmathematik, Lehrbuch der Zins-, Renten-, Tilgungs-, Kurs- und Renditerechnung, 5th Edition, Munich. [21] Kruschwitz, L. (2018): Finanzmathematik, 6th Edition, Berlin & Boston, ISBN 978-3-11-058737-1.

620

B Bibliography

[22] Langmann, J. (2009): Winkelfunktionen am Einheitskreis, https://www.mathe-online.at/lernpfade/einheitskreis/?navig=r& kapitel=3, accessed 7 December 2022. [23] Locarek-Junge, H. (1997): Finanzmathematik: Lehr- und Übungsbuch, 3rd Edition, Munich. [24] Luenberger, D.G. & Ye, Y. (2016): Linear and Nonlinear Programming, 4th Edition, Heidelberg & New York. [25] Mohr, J. (2017): Kosinus-, Sinus- und Tangenswerte, http://kilchb.de/cosinuswerte.php, accessed 7 December 2022. [26] Peren, F.W. (1986): Einkommen, Konsum und Ersparnis der privaten Haushalte in der Bundesrepublik Deutschland seit 1970: Analyse unter Verwendung makrooekonomischer Konsumfunktionen, Frankfurt am Main, ISBN 3-8204-9006-X. [27] Peren, F.W. (1990): Messung und Analyse von Substitutions- und Fortschrittseffekten in den Sektoren der westdeutschen Textilindustrie (1976 - 1986), Muenster, ISBN 3-88660-726-7. [28] Peren, F.W. (2012): The Peren Theorem, New York, unpublished manuscript. [29] Peren, F.W. (2018): Das Peren-Theorem, in: Gadatsch, A. et al. (Ed.): NachhaltigesWirtschaften im digitalen Zeitalter, Berlin, pages 419-424. [30] Peren, F.W. (2019): Unsustainable Future: The Mathematical Frame in Which We Live. In: Review of Business: Interdisciplinary Journal on Risk and Society, 39(2), pages 32-35. Peter J. Tobin College of Business at St. John’s University in New York (Ed.), 15. Juli 2019, New York. [31] Peren, F.W. (2021): Math for Business and Economics: Compendium of Essential Formulas, Berlin & Heidelberg, ISBN 978-3-662-63248-2.

B Bibliography

621

[32] Peren, F.W. (2022a): Formelsammlung Finanzmathematik, Berlin & Heidelberg, ISBN 978-3-662-65367-8. [33] Peren, F.W. (2022b): Formelsammlung Wirtschaftsmathematik, 4th Edition, Berlin & Heidelberg, ISBN 978-3-662-64835-3. [34] Peren, F.W. (2022c): Formelsammlung Wirtschaftsstatistik, 5th Edition, Berlin & Heidelberg, ISBN 978-3-662-66076-8. [35] Peren, F.W. & Akyazgan, P.D. (2021): Peren Teoremi: ˙Içinde ˘ Ya¸sadıgımız Matematiksel Çerçeve. In: Journal of Ekonomi, 3(1), pages 1-2. [36] Rathmann, H. (1990): Preismessung bei Privatkrediten von Banken und Sparkassen: Eine Analyse unter besonderer Berücksichtigung der Preisangabenverordnung. In: Hagener betriebswirtschaftliche Abhandlungen, Volume 8, Heidelberg. [37] Riebel, P. (2013): Einzelkosten- und Deckungsbeitragsrechnung. Grundfragen einer markt- und entscheidungsorientierten Unternehmensrechnung, 6th Edition, Wiesbaden. [38] Schweitzer, M.; Küpper, H.-U.; Friedl, G.; Hofmann, C.; Pedell, B. (2015): Systeme der Kosten- und Erlösrechnung, 11th Edition, Munich. [39] Serlo Education e. V. (Ed.) (2016): Ableitungen, Symmetrien und Umkehrfunktionen trigonometrischer Funktionen, https://de.serlo.org/mathe/funktionen/wichtige-funktionstypen -ihre-eigenschaften/trigonometrische-funktionen/ableitungensymmetrien-umkehrfunktionen-trigonometrischer-funktionen, accessed 19 July 2019. [40] Serlo Education e. V. (Ed.) (2018): Sinusfunktion und Kosinusfunktion, https://de.serlo.org/mathe/funktionen/wichtigefunktionstypen-ihre-eigenschaften/trigonometrische-funktionen/ sinusfunktion-kosinusfunktion, accessed 7 December 2022.

622

B Bibliography

[41] Serlo Education e. V. (Ed.) (2019): Tangensfunktion, https://de.serlo.org/mathe/funktionen/wichtige-funktionstypenihre-eigenschaften/trigonometrische-funktionen/tangensfunktion, accessed 19 July 2019. [42] Stratosphere Digital (2020): https://formula.amardesh.com/ mathematics/trigonometric-functions-of-common-angles/, accessed 4 June 2020. [43] Tucker, W.R. (2000): Effective Interest Rate (EIR). In: Bankakademie Micro Banking Competence Center (Ed.): https://web.archive.org/web/20051103034219/http: //www.uncdf.org/mfdl/readings/EIR_Tucker.pdf, accessed 13 October 2020. [44] Twin, A. (2019): Open-End Credit. In: Investopedia (Ed.): https://www.investopedia.com/terms/o/openendcredit.asp, accessed 9 December 2022. [45] Vanderbei, R.J. (2014): Linear Programming. Foundations and Extensions, 4th Edition, New York. [46] Wessels, P. (1992): Zinsrecht in Deutschland und England: eine rechtsvergleichende Untersuchung. In: Münsterische Beiträge zur Rechtwissenschaft, Volume 59, Berlin. [47] Wikimedia Foundation Inc. (Ed.) (2020): https://wikipedia.org/wiki/Annual_percentage_rate#cite_note-9, accessed 9 December 2022. [48] Wikimedia Foundation Inc. (Ed.) (2020): https://en.wikipedia.org/wiki/Population_growth, accessed 9 December 2022. [49] Wikimedia Foundation Inc. (Ed.) (2018): Areasinus hyperbolicus und Areakosinus hyperbolicus, https://de.wikipedia.org/wiki/Areasinus_hyperbolicus_und_ Areakosinus_hyperbolicus, accessed 19 July 2019.

B Bibliography

623

[50] Wikimedia Foundation Inc. (Ed.) (2018): Areatangens hyperbolicus und Areakotangens hyperbolicus, https://de.wikipedia.org/wiki/Areatangens_hyperbolicus_und_ Areakotangens_hyperbolicus, accessed 19 July 2019. [51] Wikimedia Foundation Inc. (Ed.) (2018): Arkussinus und Arkuskosinus, https://de.wikipedia.org/wiki/Arkussinus_und_Arkuskosinus, accessed 7 December 2022. [52] Wikimedia Foundation Inc. (Ed.) (2018): Arkustangens und Arkuskotangens, https://de.wikipedia.org/wiki/Arkustangens_und_Arkuskotangens, accessed 7 December 2022. [53] Wikimedia Foundation Inc. (Ed.) (2018): Formelsammlung Trigonometrie, https://de.wikipedia.org/wiki/Formelsammlung_Trigonometrie, accessed 7 December 2022. [54] Wikimedia Foundation Inc. (Ed.) (2018): Sinus hyperbolicus und Kosinus hyperbolicus, https://de.wikipedia.org/wiki/Sinus_hyperbolicus_und_Kosinus_ hyperbolicus, accessed 19 July 2019. [55] Wikimedia Foundation Inc. (Ed.) (2018): Tangens hyperbolicus und Kotangens hyperbolicus, https://de.wikipedia.org/wiki/Tangens_hyperbolicus_und_ Kotangens_hyperbolicus, accessed 19 July 2019. [56] Wikimedia Foundation Inc. (Ed.) (2018): Tangens und Kotangens, https://de.wikipedia.org/wiki/Tangens_und_Kotangens, accessed 7 December 2022. [57] Wikimedia Foundation Inc. (Ed.) (2019): Sekans hyperbolicus und Kosekans hyperbolicus, https://de.wikipedia.org/wiki/ Sekans_hyperbolicus_und_Kosekans_hyperbolicus, accessed 7 December 2022.

624

B Bibliography

[58] Wikimedia Foundation Inc. (Ed.) (2019): Sinus und Kosinus, https://de.wikipedia.org/wiki/Sinus_und_Kosinus, accessed 7 December 2022. [59] World Wide Fund For Nature - WWF (Ed.) (2017): http://wwf.panda.org/about_our_Earth/all_publications/living_ planet_report_timeline/lpr_2012/, accessed 9 December 2022. [60] Yildirim, M. (2019): Trigonometrie, https://www.schulminator.com/mathematik/trigonometrie, accessed 7 December 2022. [61] Zimmermann, H.-J. (2005): Operations Research. Methoden und Modelle. Für Wirtschaftsingenieure, Betriebswirte und Informatiker, Wiesbaden.

Index

Symbols a − b − c Formula. . . . . . . . . . . .62 A Absolute Value . . . . . . . . . . . . . 26 Addition Method . . . . . . . . . . . . 60 Adjoint of a Matrix . . . . . . . . . 125 Determination of the Inverse with the Usage of the Adjoint . . . . . . . . . . . . 127 Algebra . . . . . . . . . . . . . . . . . . . . . 51 Amount of Annuity Final Annuity Value Factor. . . . . . . . . . . . . . . . . . .181 Annual Percentage Rate . . . 168 Annuity Calculation . . . . . . . . 180 Annual Annuity with Sub-Annual Interest . . . . . 187 Annuity Factor . . . . . . . . . . 182 Finite, Regular Annuity . . 183 Finite, Variable Annuity . . 213 Irregular Annuity . . . . . . . . 213 Perpetuity . . . . . . . . . . . . . . . 234 Sub-Annual Annuity with Annual Interest . . . . . . . . . . 190 Sub-Annual Annuity with Sub-Annual Interest . . . . . 194 Annuity Factor . . . . . . . . . . . . . 182 Annuity Method. . . . . . . . . . . .290 Annuity Repayment. . . . . . . .238 Approximation Methods . . . . . 75 Arc Elasticity . . . . . . . . . . . . . . 486 Arcus Function . . . . . . . . . . . . 369 Arcus Functions of Negative x-values . . . . . . . 370

Area Functions . . . . . . . . . . . . 379 Arithmetic . . . . . . . . . . . . . . . . . . 15 Arithmetic Relations and Links. . . . . . . . . . . . . . . . . . . . . . . . .1 Arithmetic Sequences . . . . . . 46 Asymptotes . . . . . . . . . . . . . . . 404 Asymptotic Curve . . . . . . . 404 Horizontal . . . . . . . . . . . . . . . 405 Oblique . . . . . . . . . . . . . . . . . 408 Vertical . . . . . . . . . . . . . . . . . 407 Axioms . . . . . . . . . . . . . . . . . . . . . 25 B Bijection . . . . . . . . . . . . . . . . . . . 327 Binomial Coefficient . . . . . . . . 40 Binomial Formulas . . . . . . . . . . 30 Binomial Theorem . . . . . . . . . . 31 for Natural Exponents . . . . 31 for Real Exponents . . . . . . . 31 Binomials . . . . . . . . . . . . . . . . . 378 Biquadratic Equations . . . . . . 67 Boundedness . . . . . . . . . . . . . 382 Bounds Theorem for Differential Calculus . . . . . . . 458 Buyer’s Market and Seller’s Market . . . . . . . . . . . . 508 C Calculation of Interest.141, 142 Annual Interest . . . . . . . . . . 142 Composite Interest . . . . . . 146 Compound Computation of Interest . . . . . . . . . . . . . . . 144 Interest . . . . . . . . . . . . . . . . . 141 Interest Factor . . . . . . . . . . 142

© Springer-Verlag GmbH Germany, part of Springer Nature 2023 F. W. Peren, Math for Business and Economics, https://doi.org/10.1007/978-3-662-66975-4

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626 Interest Period . . . . . . . . . . 142 Interest Rate . . . . . . . . . . . . 141 Classification of Functions . 333 Combinations . . . . . . . . . . . . . 136 Combinatorics . . . . . . . . . . . . . 129 Completing the Square . . . . . 63 Composition . . . . . . . . . . . . . . . 329 Concavity and Convexity. . .402 Continuity . . . . . . . . . . . . . . . . . 394 Conversions of Terms . . . . . . . 30 Cost Function Cost Function According to the Law of Diminishing Returns . . . . . . . . . . . . . . . . . 533 Multi-dimensional Cost Allocation Principles . . . . 549 One-dimensional Cost Allocation Principles . . . . 549 Cost Functions . . . . . . . . . . . . 517 Cournot Point . . . . . . . . . . . . . 557 Cubic Equations with One Variable . . . . . . . . . . . . . . . 65 Cubic Equations without Absolute Term . . . . . . . . . . . . . . 66 Curve Sketching . . . . . . . . . . . 425 Cyclometric Functions . . . . . 369 D Decimal System, Decadic System . . . . . . . . . . . . . . . . . . . . . 23 Definite Integral . . . . . . . . . . . 464 Demand Function/Inverse Demand Function . . . . . . . . . 505 Demand Gap . . . . . . . . . . . . . . 509 Depreciation. . . . . . . . . . . . . . .173 Arithmetic-Degressive . . . 174 Extraordinary . . . . . . . . . . . 179 Geometric-Degressive . . 176 Linear. . . . . . . . . . . . . . . . . . .173 Time Depreciation . . . . . . 173

Index Units of Production Depreciation . . . . . . . . . . . . 178 Derivation Rules . . . . . . . . . . . 422 Higher Derivations . . . . . . 424 Derivative Function . . . . . . . . 419 Derivatives of Elementary Functions . . . . . . . . . . . . . . . . . 420 Difference Quotient . . . . . . . . 417 Differential Calculus . . . . . . . 417 Differential Quotient . . . . . . . 418 Differentials . . . . . . . . . . . . . . . 453 Differentiation of Functions with More Than One Independent Variable . . . . . . . . . . 435 Differentiation of Functions with Parameters . . . . . . . . . . . 425 Direct Costs . . . . . . . . . . . . . . . 546 Dual Simplex Algorithm . . . . 315 Dual System (Binary System) . . . . . . . . . . . . . . . . . . . . 23 E Economic Functions . . . . . . . 503 Elasticities. . . . . . . . . . . . . . . . .485 Arc Elasticities . . . . . . . . . . 486 Cross Elasticity of Demand . . . . . . . . . . . . . . . . 499 Point Elasticity . . . . . . . . . . 491 Elementary Calculus . . . . . . . . 24 Elementary Foundations . . . . 24 Equalisation Method . . . . . . . . 59 Equations . . . . . . . . . . . . . . . . . . 51 Equivalent Transformations of Equations . . . . . . . . 52 Exponential Equations. . . .71 Fractional Equations. . . . . .53 Linear Equations . . . . . . . . . 53 Logarithmic Equations . . . . 73 Non-linear Equations . . . . . 62 Radical Equations . . . . . . . . 69

Index Transcendental Equations 71 Universal Equations . . . . . . 52 Equation of the nth Degree . . . . . . . . . . . . . . . 68 Equivalent Transformations of Equations . . . . . . . . . . . . . . . . 52 Exponential Functions . . . . . . . . . . . . . 342, 348 Reflections . . . . . . . . . . . . . . 345 Shift . . . . . . . . . . . . . . . . . . . . 347 Stretch/Compression . . . . 344 Exponential Functions . . . . . . . . . . . . . . . . . . . . 4 Extremes . . . . . . . . . . . . . 358, 400 F Factorial . . . . . . . . . . . . . . . . . . . . 39 Factorisation . . . . . . . . . . . . . . . . 25 Falk’s Scheme . . . . . . . . . . . . 106 Financial Mathematics . . . . . 141 Fixed Costs . . . . . . . . . . . . . . . 518 Fractional Equations . . . . . . . . 53 Fractions . . . . . . . . . . . . . . . . . . . 27 Functions . . . . . . . . . . . . . . . 7, 325 Broken Rational Functions . . . . . . . . . . 333, 334 Classification. . . . . . . . . . . .333 Composition . . . . . . . . . . . . 329 Exponential Functions. . .342 Non-rational Functions . . . . . . . . . . . . . . . 333, 338 Polynomial Functions . . . . . . . . . . . . . . . 333, 334 Rational Functions . 333, 334 Root Function . . . . . . . . . . . 341 Transcendental Functions . . . . . 333, 338, 342 Fundamental Arithmetic Operations . . . . . . . . . . . . . . . . . 24

627 Fundamentals of Financial Mathematics . . . . . . . . . . 283 G General Approximation Method (Fixed-point Iteration) . . . . . . . . . . . . . . . . . . . . 80 Geometric Sequence . . . . . . . 46 Goniometric Transformations . . . . . . . . . . . . . . . . . . . . . . 364 Greek Alphabet . . . . . . . . . . . . . . 9 H Hessian matrix . . . . . . . . . . . . 440 Homogeneity . . . . . . . . . . . . . . 398 Horner’s Scheme (Horner’s Method) . . . . . . . . . . . . . . . . . . . . 29 Hyperbolic Functions . . . . . . . . . . . . . . . 4, 333, 371 I Improper Limit . . . . . . . . . . . . . . 45 Inclusion . . . . . . . . . . . . . . . . . . . . 16 Income Elasticity of Demand . . . . . . . . . . . . . . . . . . . 501 Indefinite Integral . . . . . . . . . . 460 Indirect Costs . . . . . . . . . . . . . 546 Inequations . . . . . . . . . . . . . . . . . 51 Fractional Inequations with One Variable . . . . . . . . 54 Linear Inequations with Multiple Variables . . . . 61 Linear Inequations with One Variable . . . . . . . . 56 Infimum . . . . . . . . . . . . . . . . . 16, 43 Infinite Discontinuity . . . . . . . 394 Inflection Points . . . . . . . . . . . 402 Injection . . . . . . . . . . . . . . . . . . . 327 Integral Calculus . . . . . . . . . . 459

628 Antiderivative. . . . . . .459, 460 Definite Integral . . . . . . . . . 464 Elementary Calculation Rules for the Indefinite Integral . . . . . . . . . . . . . . . . . 463 Indefinite Integral. . . . . . . .460 Integration by Substitution . . . . . . . . . . . . . . . . . . . . . 475 Multiple Integrals . . . . . . . . 476 Partial Integration . . . . . . . 473 Relationship between the Definite and the Indefinite Integral . . . . . . . 468 Special Techniques of Integration . . . . . . . . . . . . . . 473 Interest Calculation Compound Interest . 160, 161 Interest During the Period . . . . . . . . . . . . . . . . . . 158 Interest Factor. . . . . .180, 231 Mixed Interest . . . . . . . . . . . 162 Nominal Interest Rate . . . 159 Period . . . . . . . . . . . . . . . . . . 143 Steady Interest Rate . . . . 163 Inverse Function . . . . . . . . . . . 331 Inverse of a Matrix Determination of the Inverse with the Usage of the Gaussian Elimination Method . . . . . . . . . . . . . 109 Investment Calculation . . . . 279 Amortisation Calculation 286 Amount of Capital Method . . . . . . . . . . . . . . . . . 287 Annuity Method . . . . . . . . . 290 Cost Comparison Method . . . . . . . . . . . . . . . . . 286 Final Asset Value Method . . . . . . . . . . . . . . . . . 287 Internal Rate of Return Method . . . . . . . . . . . . . . . . . 293

Index Methods of Dynamic Investment Calculation . . 286 Methods of Static Investment Calculation . . 286 Net Present Value Method . . . . . . . . . . . . . . . . . 287 Pay-Back Method . . . . . . . 286 Pay-Off Method . . . . . . . . . 286 Pay-Out Method. . . . . . . . .286 Profit Comparison Method . . . . . . . . . . . . . . . . . 286 Profitability Calculation . . 286 J Jump Discontinuity . . . . . . . . 397 L Lagrange Method. . . . . . . . . .297 Limit . . . . . . . . . . . . . . . . . . . . . . . . . 3 Limit of a Sequence . . . . . . . . 44 Linear Algebra . . . . . . . . . . . . . . 87 Linear Equations. . . . . . . . . . . .53 with Multiple Variables . . . . 56 with One Variable . . . . . . . . 53 Linear Optimisation . . . . . . . . 302 Graphical Solution . . . . . . 303 Linear Optimisation (Linear Programming Approach) . . . . . . . . . . . . . . . . . 297 Establishing the Linear Programming Approach . 303 Local Extremes . . . . . . . . . . . . 400 Logarithm . . . . . . . . . . . . . . . . . . . . 4 Logarithmic Functions . . . . . . . . 330, 333, 348 Reflection . . . . . . . . . . . . . . . 350 Shift . . . . . . . . . . . . . . . . . . . . 352 Stretch/Compression . . . . 353 Logarithmic Equations . . . . . . 73

Index Logarithms . . . . . . . . . . . . . . . . . 37 Common Logarithm . . . . . . 38 Logarithm to an Arbitrary Base . . . . . . . . . . . . . . . . . . . . . 39 Logarithmic Laws . . . . . . . . 38 Logarithmic Systems . . . . . 38 Natural Logarithm . . . . . . . . 39 Logic . . . . . . . . . . . . . . . . . . . . . . . 11 Mathematical Logic. . . . . . .11 Propositional Logic . . . . . . . 11 M Marginal Cost Function . . . . 521 Market Equilibrium . . . . . . . . 507 Mathematical Signs and Symbols . . . . . . . . . . . . . . . . . . . . . 1 Matrices . . . . . . . . . . . . . . . . . . 5, 87 Addition . . . . . . . . . . . . . . . . . . 94 Adjoint of a Matrix . . . . . . . 125 Determinant of a Matrix . 117 Equality/Inequality . . . . . . . . 88 Inverse of a Matrix . . . . . . 107 Multiplication . . . . . . . . . . . . . 96 Multiplication of a Matrix by a Column Vector . . . . 100 Multiplication of a Matrix with a Scalar . . . . . . . . . . . . . 96 Multiplication of a Row Vector by a Matrix . . . . . 102 Multiplication of Two Matrices . . . . . . . . . . . . . . . . 103 Operations with Matrices 94 Rank of a Matrix . . . . . . . . 113 Special Matrices . . . . . . . . . 92 Transposed Matrix . . . . . . . . 89 Mean Value Theorem for Differential Calculus . . . . . . . 455 Generalized Mean Value Theorem . . . . . . . . . 456 Minor . . . . . . . . . . . . . . . . . . . . . . 117

629 Monotonicity. . . . . . . . . . . . . . .401 Multiple Integrals . . . . . . . . . . 476 N Neoclassical Cost Function . . . . . . . . . . . . . . . . . . . . . . . . 525 Newton’s Method (Tangent Method) . . . . . . . . . . . . . . . . . . . . 77 Non-linear Equations . . . . . . . 62 Normal Lines to a Curve . . . 411 Null Sequence . . . . . . . . . . . . . . 45 Numeral Systems . . . . . . . . . . . 22 O Operations with Matrices. . . .94 Optimisation of Linear Models . . . . . . . . . . . . . . . . . . . . 297 Order Structures . . . . . . . . . . . . . 7 P p/q Formula . . . . . . . . . . . . . . . . 62 Partial Derivatives . . . . . . . . . 435 1st Order . . . . . . . . . . . . . . . . 435 2nd Order . . . . . . . . . . . . . . . 438 Local Extrema . . . . . . . . . . 440 rth Order . . . . . . . . . . . . . . . . 439 Relative Extrema with m Constraints . . . . . . . . . . . 449 Schwarz’ Theorem . . . . . . 438 Partial Differential . . . . . . . . . 453 Partial Integration. . . . . . . . . .473 Percentage Annuity . . . . . . . . 257 Peren Theorem . . . . . . . . . . . . 563 Periodicity . . . . . . . . . . . . 363, 399 Permutations . . . . . . . . . . . . . . 133 Point Elasticity . . . . . . . . . . . . . 491 Pole . . . . . . . . . . . . . . . . . . 336, 337 Polynomial Division . . . . . . . . . 27 Power Functions. . . . . . . . . . .333

630 Powers . . . . . . . . . . . . . . . . . . . . . 34 Pragmatic Signs . . . . . . . . . . . . . 1 Present Value Annuity Present Value Factor. . . . . . . . . . . . . . . . . . .181 Present Value of Annuity . . 180 Price Elasticity of Demand . 494 Primal Simplex Algorithm . . 308 Product Notation . . . . . . . . . . . . 33 Profit Function . . . . . . . . . . . . . 554 Propositional Variable . . . . . . . 11 Q Quadratic Equations . . . . . . . . 62 R Radical Equations . . . . . . . . . . 69 Rate of Return Method (internal) . . . . . . . . . . . . . . . . . . 293 Real Functions . . . . . . . 333, 382 Regula falsi (Secant Method) . . . . . . . . . . . . . . . . . . . . 75 Relations. . . . . . . . . . . . . . .1, 7, 26 Removable Discontinuity . . 396 Repayment with Discount (Disagio) . . 249 Repayment by Instalments.241 Revenue Function . . . . . . . . . 511 Roman Numeral System . . . . 24 Root Function. . . .333, 338, 341 Roots . . . . . . . . . . . . . . . . . . . . . . . 34 Rules of Calculation for the Multiplication of Matrices . . 104 S Scalar . . . . . . . . . . . . . . . . . . . . . . 96 Scalar Product . . . . . . . . . . . . . . 98 Schwarz’ Theorem . . . . . . . . 438 Secant Method . . . . . . . . . . . . . 75

Index Sequences . . . . . . . . . . . . . . . . . 41 Series . . . . . . . . . . . . . . . . . . . . . . 47 Arithmetic Series . . . . . . . . . 47 Geometric Series. . . . . . . . .48 Infinite Geometric Series . 49 Sets . . . . . . . . . . . . . . . . . . . . . . 6, 15 Bounds of a Set . . . . . . . . . . 16 Complement of the Set . . . 18 Intersection of Two Sets . . 17 Intervals. . . . . . . . . . . . . . . . . .21 Laws . . . . . . . . . . . . . . . . . . . . . 19 Limits of a Set . . . . . . . . . . . . 16 Power Set . . . . . . . . . . . . . . . . 18 Product of Two Sets . . . . . . 18 Relations . . . . . . . . . . . . . . . . . 19 Relative Complement of Two Sets . . . . . . . . . . . . . . . . . 17 Rules of Calculation . . . . . . 19 Set Operations . . . . . . . . . . . 17 Set Relations . . . . . . . . . . . . . 16 Sets of Numbers . . . . . . . . . . . . . 2 Shift . . . . . . . . . . . . . . . . . . . . . . . 389 Signum . . . . . . . . . . . . . . . . . . . . . 26 Sinking Fund Calculation . . 235 Annuity Repayment . . . . . 238 Grace Periods . . . . . . . . . . 255 Repayment by Instalments . . . . . . . . . . . . . . . . . . . 241 Repayment During the Year . . . . . . . . . . . . . . . . . . . . 266 Repayment of Bonds . . . . 260 Repayment with Premium . . . . . . . . . . . . . . . . . . . 243 Rounded Annuities . . . . . . 257 Special Numbers and Links . . 3 Special Techniques of Integration . . . . . . . . . . . . . . . . . . . . 473 Substitution Method . . . . . . . . 58 Summation Notation . . . . . . . . 32 Supply Function . . . . . . . . . . . 503 Supply Gap . . . . . . . . . . . . . . . 509

Index

631

Supremum . . . . . . . . . . . . . . 16, 43 Surjection . . . . . . . . . . . . . . . . . 327 Symmetry . . . . . . . . . . . . . . . . . 384 Axial Symmetry . . . . . . . . . 384 Point Symmetry . . . . . . . . . 386 Systems of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . 57

Trigonometric Functions . 4, 354 Truth Tables . . . . . . . . . . . . . . . . 12

T

Variable Costs . . . . . . . . . . . . . 519 Variations . . . . . . . . . . . . . . . . . 135 Vectors . . . . . . . . . . . . . . . 5, 91, 92 Scalar Product of Two Vectors . . . . . . . . . . . . . . 98 Special Vectors. . . . . . . . . . .92 Vertex Form . . . . . . . . . . . . . . . 391

Tangent Lines to a Curve . . 410 Tangent Plane . . . . . . . . . . . . . 440 Terms . . . . . . . . . . . . . . . . . . . . . . 30 Polynomial Terms . . . . . . . . 32 Theorems of Differentiable Functions . . . . . . . . . . . . . . . . . 455 Total Differential . . . . . . . . . . . 454 Transcendental Equations . . 71 Transformation . . . . . . . . . . . . 389

U Universal Equations . . . . . . . . 52 V

Z Zero . . . . . . . . . . . . . . . . . . . . . . . 399